Idoc.vn luong giac-ly-thuyet-bai-tap-co-loi-giai

CHÖÔNG 1: COÂNG THÖÙC LÖÔÏNG GIAÙC
I. Ñònh nghóa
Treân maët phaúng Oxy cho ñöôøng troøn löôïng giaùc taâm O baùn kính R=1 vaø ñieåm M
treân ñöôøng troøn löôïng giaùc maø sñ AM = β vôùi 0 2≤ β ≤ π
Ñaët k2 ,k Zα = β + π ∈
Ta ñònh nghóa:
sin OKα =
cos OHα =
sin
tg
cos
α
α =
α
vôùi cos 0α ≠
cos
cotg
sin
α
α =
α
vôùi sin 0α ≠
II. Baûng giaù trò löôïng giaùc cuûa moät soá cung (hay goùc) ñaëc bieät
Goùc α
Giaù trò
( )o
0 0
( )o
30
6
π
( )o
45
4
π
( )o
60
3
π
( )o
90
2
π
sinα 0 1
2
2
2
3
2
1
cosα 1 3
2
2
2
1
2
0
tgα 0 3
3
1 3 ||
cotgα || 3 1 3
3
0
III. Heä thöùc cô baûn
2 2
sin cos 1α + α =
2
2
1
1 tg
cos
+ α =
α
vôùi ( )k k Z
2
π
α ≠ + π ∈
2
2
1
t cotg
sin
+ =
α
vôùi ( )k k Zα ≠ π ∈
IV. Cung lieân keát (Caùch nhôù: cos ñoái, sin buø, tang sai π ; phuï cheùo)
a. Ñoái nhau: α vaø −α
( )sin sin−α = − α
( )cos cos−α = α
( ) ( )tg tg−α = − α
( ) ( )cotg cotg−α = − α
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b. Buø nhau: α vaø π − α
( )
( )
( )
( )
sin sin
cos cos
tg tg
cotg cotg
π − α = α
π − α = − α
π − α = − α
π − α = − α
c. Sai nhau π : vaøα π + α
( )
( )
( )
( )
sin sin
cos cos
tg tg
cotg cotg
π + α = − α
π + α = − α
π + α = α
π + α = α
d. Phuï nhau: α vaø
2
π
− α
sin cos
2
cos sin
2
tg cotg
2
cotg tg
2
π⎛ ⎞
− α = α⎜ ⎟
⎝ ⎠
π⎛ ⎞
− α = α⎜ ⎟
⎝ ⎠
π⎛ ⎞
− α = α⎜ ⎟
⎝ ⎠
π⎛ ⎞
− α = α⎜ ⎟
⎝ ⎠
e.Sai nhau
2
π
: vaøα
2
π
+ α
sin cos
2
cos sin
2
tg cotg
2
cotg tg
2
π⎛ ⎞
+ α = α⎜ ⎟
⎝ ⎠
π⎛ ⎞
+ α = − α⎜ ⎟
⎝ ⎠
π⎛ ⎞
+ α = − α⎜ ⎟
⎝ ⎠
π⎛ ⎞
+ α = − α⎜ ⎟
⎝ ⎠
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f.
( ) ( )
( ) ( )
( )
( )
+ π = − ∈
+ π = − ∈
+ π = ∈
+ π =
k
k
sin x k 1 sinx,k Z
cos x k 1 cosx,k Z
tg x k tgx,k Z
cotg x k cotgx
V. Coâng thöùc coäng
( )
( )
( )
sin a b sinacosb sinbcosa
cos a b cosacosb sinasinb
tga tgb
tg a b
1 tgatgb
± = ±
± =
±
± =
∓
∓
VI. Coâng thöùc nhaân ñoâi
=
= − = − =
=
−
−
=
2 2 2 2
2
2
sin2a 2sinacosa
cos2a cos a sin a 1 2sin a 2cos a 1
2tga
tg2a
1 tg a
cotg a 1
cotg2a
2cotga
−
VII. Coâng thöùc nhaân ba:
3
3
sin3a 3sina 4sin a
cos3a 4cos a 3cosa
= −
= −
VIII. Coâng thöùc haï baäc:
( )
( )
2
2
2
1
sin a 1 cos2a
2
1
cos a 1 cos2a
2
1 cos2a
tg a
1 cos2a
= −
= +
−
=
+
IX. Coâng thöùc chia ñoâi
Ñaët
a
t tg
2
= (vôùi )a k2≠ π + π
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2
2
2
2
2t
sina
1 t
1 t
cosa
1 t
2t
tga
1 t
=
+
−
=
+
=
−
X. Coâng thöùc bieán ñoåi toång thaønh tích
( )
( )
a b a b
cosa cosb 2cos cos
2 2
a b a b
cosa cosb 2sin sin
2 2
a b a b
sina sinb 2cos sin
2 2
a b a b
sina sinb 2cos sin
2 2
sin a b
tga tgb
cosacosb
sin b a
cotga cotgb
sina.sinb
+ −
+ =
+ −
− = −
+ −
+ =
+ −
− =
±
± =
±
± =
XI. Coâng thöùc bieån ñoåi tích thaønh toång
( ) ( )
( ) ( )
( ) ( )
1
cosa.cosb cos a b cos a b
2
1
sina.sinb cos a b cos a b
2
1
sina.cosb sin a b sin a b
2
= ⎡ + + −⎣ ⎦
−
⎤
= ⎡ + − −⎣ ⎦⎤
= ⎡ + + − ⎤⎣ ⎦
Baøi 1: Chöùng minh
4 4
6 6
sin a cos a 1 2
sin a cos a 1 3
+ −
=
+ −
Ta coù:
( )
24 4 2 2 2 2 2
sin a cos a 1 sin a cos a 2sin acos a 1 2sin acos a+ − = + − − = − 2
Vaø:
( )( )
( )
6 6 2 2 4 2 2 4
4 4 2 2
2 2 2 2
2 2
sin a cos a 1 sin a cos a sin a sin acos a cos a 1
sin a cos a sin acos a 1
1 2sin acos a sin acos a 1
3sin acos a
+ − = + − +
= + − −
= − − −
= −
−
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Do ñoù:
4 4 2 2
6 6 2 2
sin a cos a 1 2sin acos a 2
sin a cos a 1 3sin acos a 3
+ − −
= =
+ − −
Baøi 2: Ruùt goïn bieåu thöùc
( )
2
2
1 cosx1 cosx
A 1
sinx sin x
⎡ ⎤−+
= = +⎢ ⎥
⎢ ⎥⎣ ⎦
Tính giaù trò A neáu
1
cosx
2
= − vaø x
2
π
< < π
Ta coù:
2 2
2
1 cosx sin x 1 2cosx cos x
A
sinx sin x
⎛ ⎞+ + − +
= ⎜ ⎟
⎝ ⎠
( )
2
2 1 cosx1 cosx
A .
sinx sin x
−+
⇔ =
( )2 2
3 3
2 1 cos x 2sin x 2
A
sin x sin x sinx
−
⇔ = = = (vôùi sinx 0≠ )
Ta coù: 2 2 1 3
sin x 1 cos x 1
4 4
= − = − =
Do: x
2
π
< < π neân sinx 0>
Vaäy
3
sinx
2
=
Do ñoù
2 4 4
A
sinx 33
= = =
3
Baøi 3: Chöùng minh caùc bieåu thöùc sau ñaây khoâng phuï thuoäc x:
a. A = −4 4 2 2 2
2cos x sin x sin xcos x 3sin x+ +
b.
2 cotgx 1
tgx 1 cotgx 1
+
− −
B = +
a. Ta coù:
4 4 2 2
A 2cos x sin x sin xcos x 3sin x= − + + 2
( ) ( ) ( )
( )
24 2 2 2 2
4 2 4 2 4
A 2cos x 1 cos x 1 cos x cos x 3 1 cos x
A 2cos x 1 2cos x cos x cos x cos x 3 3cos x
⇔ = − − + − + −
⇔ = − − + + − + − 2
A 2⇔ = (khoâng phuï thuoäc x)
b. Vôùi ñieàu kieän sinx.cosx 0,tgx 1≠ ≠
Ta coù:
2 cotgx
B
tgx 1 cotgx 1
1+
= +
− −
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1
1
2 2tgx
B
1tgx 1 tgx 1 1 tgx1
tgx
+
+
⇔ = + = +
− −−
1 tgx
−
( )2 1 tgx 1 tgx
B 1
tgx 1 tgx 1
− − −
⇔ = = = −
− −
(khoâng phuï thuoäc vaøo x)
Baøi 4: Chöùng minh
( )
2 2 2
2 2
2 2 2
1 cosa1 cosa cos b sin c
1 cotg bcotg c cotga 1
2sina sin a sin bsin c
⎡ ⎤−+ −
− + − =⎢ ⎥
⎢ ⎥⎣ ⎦
−
Ta coù:
*
2 2
2 2
2 2
cos b sin c
cotg b.cotg c
sin b.sin c
−
−
2
2 2
2 2
cotg b 1
cotg bcotg c
sin c sin b
= − −
( ) ( )2 2 2 2 2
cotg b 1 cotg c 1 cotg b cotg bcotg c 1= + − + − = − (1)
*
( )
2
2
1 cosa1 cosa
1
2sina sin a
⎡ ⎤−+
−⎢ ⎥
⎢ ⎥⎣ ⎦
( )
2
2
1 cosa1 cosa
1
2sina 1 cos a
⎡ ⎤−+
= −⎢ ⎥
−⎢ ⎥⎣ ⎦
1 cosa 1 cosa
1
2sina 1 cosa
+ −⎡ ⎤
= −⎢ ⎥+⎣ ⎦
1 cosa 2cosa
. c
2sina 1 cosa
+
= =
+
otga (2)
Laáy (1) + (2) ta ñöôïc ñieàu phaûi chöùng minh xong.
Baøi 5: Cho tuøy yù vôùi ba goùc ñeàu laø nhoïn.ABCΔ
Tìm giaù trò nhoû nhaát cuûa P tgA.tgB.tgC=
Ta coù: A B C+ = π −
Neân: ( )tg A B tgC+ = −
tgA tgB
tgC
1 tgA.tgB
+
⇔ =
−
−
+
tgA tgB tgC tgA.tgB.tgC⇔ + = − +
Vaäy: P tgA.tgB.tgC tgA tgB tgC= = +
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AÙp duïng baát ñaúng thöùc Cauchy cho ba soá döông tg ta ñöôïcA,tgB,tgC
3tgA tgB tgC 3 tgA.tgB.tgC+ + ≥
3
P 3 P⇔ ≥
3 2
P 3
P 3 3
⇔ ≥
⇔ ≥
Daáu “=” xaûy ra
= =⎧
π⎪
⇔ ⇔ =⎨ π
< <⎪⎩
tgA tgB tgC
A B C
30 A,B,C
2
= =
Do ñoù: MinP 3 3 A B C
3
π
= ⇔ = = =
Baøi 6 : Tìm giaù trò lôùn nhaát vaø nhoû nhaát cuûa
a/ 8 4
y 2sin x cos 2x= +
b/ 4
y sin x cos= − x
a/ Ta coù :
4
41 cos2x
y 2 cos 2x
2
−⎛ ⎞
= +⎜ ⎟
⎝ ⎠
Ñaët vôùi thìt cos2x= 1 t 1− ≤ ≤
( )
4 41
y 1 t
8
= − + t
=> ( )
3 31
y ' 1 t 4t
2
= − − +
Ta coù : ( )y' 0=
3 3
1 t 8t− =
⇔ 1 t 2t− =
⇔
1
t
3
=
Ta coù y(1) = 1; y(-1) = 3;
1 1
y
3 2
⎛ ⎞
=⎜ ⎟
⎝ ⎠ 7
Do ñoù : vaø
∈
=
x
y 3Max
∈
=
x
1
yMin
27
b/ Do ñieàu kieän : sin vaø co neân mieàn xaùc ñònhx 0≥ s x 0≥
π⎡ ⎤
= π + π⎢ ⎥⎣ ⎦
D k2 , k2
2
vôùi ∈k
Ñaët t cos= x xvôùi thì0 t 1≤ ≤ 4 2 2
t cos x 1 sin= = −
Neân 4
sin x 1 t= −
Vaäy
8 4
y 1 t= − − t treân [ ]D' 0,1=
Thì
( )
−
= − <
−
3
748
t
y ' 1 0
2. 1 t
[ )∀ ∈t 0; 1
Neân y giaûm treân [ 0, 1 ]. Vaäy : ( )∈
= =
x D
max y y 0 1, ( )∈
= = −
x D
min y y 1 1
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Baøi 7: Cho haøm soá 4 4
y sin x cos x 2msin x cos= + − x
Tìm giaù trò m ñeå y xaùc ñònh vôùi moïi x
Xeùt 4 4
f(x) sin x cos x 2msin x cos x= + −
( ) ( )
22 2 2
f x sin x cos x msin 2x 2sin x cos x= + − − 2
( ) 21
f x 1 sin 2x msin2x
2
= − −
Ñaët : vôùit sin 2x= [ ]t 1,∈ − 1
y xaùc ñònh x∀ ⇔ ( )f x 0 x R≥ ∀ ∈
⇔ 21
1 t [ ]mt
2
− − ≥ 0 t 1,1−∀ ∈
⇔ ( ) 2
g t t 2mt 2 0= + − ≤ [ ]t 1,1−
t
∀ ∈
Do ∀ neân g(t) coù 2 nghieäm phaân bieät t2
' m 2 0Δ = + > m 1, t2
Luùc ñoù t t1 t2
g(t) + 0 - 0
Do ñoù : yeâu caàu baøi toaùn ⇔ 1 2t 1 1≤ − < ≤
⇔ ⇔
( )
( )
1g 1 0
1g 1 0
− ≤⎧⎪
⎨
≤⎪⎩
2m 1 0
2m 1 0
− − ≤⎧
⎨
− ≤⎩
⇔
1
m
2
1
m
2
−⎧
≥⎪⎪
⎨
⎪ ≤
⎪⎩
⇔
1 1
m
2 2
− ≤ ≤
Caùch khaùc :
g t( ) 2
t 2mt 2 0= + − ≤ [ ]t 1,∀ ∈ − 1
{ }[ , ]
max ( ) max ( ), ( )
t
g t g g
∈ −
⇔ ≤ ⇔ − ≤
11
0 1 1 0
{ }max ), )m m⇔ − − − + ≤2 1 2 1 0 ⇔
1
m
2
1
m
2
−⎧
≥⎪⎪
⎨
⎪ ≤
⎪⎩
m⇔ − ≤ ≤
1 1
2 2
Baøi 8 : Chöùng minh 4 4 4 43 5 7
A sin sin sin sin
16 16 16 16 2
π π π π
= + + +
3
=
Ta coù :
7
sin sin cos
16 2 16 16
π π π π⎛ ⎞
= − =⎜ ⎟
⎝ ⎠
π π π⎛ ⎞
= − =⎜ ⎟
⎝ ⎠
5 5
sin cos cos
16 2 16 16
π3
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Maët khaùc : ( )
24 4 2 2 2
sin cos sin cos 2sin cosα + α = α + α − α α2
2 2
1 2sin cos= − α α
21
1 sin 2
2
= − α
Do ñoù : 4 4 4 47 3
A sin sin sin sin
16 16 16 16
π π π π
= + + +
5
4 4 4 43 3
sin cos sin cos
16 16 16 16
π π π⎛ ⎞ ⎛
= + + +⎜ ⎟ ⎜
⎝ ⎠ ⎝
π ⎞
⎟
⎠
2 21 1
1 sin 1 sin
2 8 2 8
3π π⎛ ⎞ ⎛
= − + −⎜ ⎟ ⎜
⎝ ⎠ ⎝
⎞
⎟
⎠
2 21 3
2 sin sin
2 8 8
π π⎛ ⎞
= − +⎜ ⎟
⎝ ⎠
2 21
2 sin cos
2 8 8
π π⎛ ⎞
= − +⎜ ⎟
⎝ ⎠
π π
=
⎝ ⎠
3
do sin cos
8 8
⎛ ⎞
⎜ ⎟
1 3
2
2 2
= − =
Baøi 9 : Chöùng minh : o o o o
16sin10 .sin 30 .sin 50 .sin70 1=
Ta coù :
o
o
A cos10 1
A
cos10 cos10
= = o
(16sin10o
cos10o
)sin30o
.sin50o
.sin70o
⇔ ( )o o
o
1 1 o
A 8sin 20 cos 40 .cos 20
2cos10
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⇔ ( )0 o
o
1 o
A 4 sin 20 cos 20 .cos 40
cos10
=
⇔ ( )o o
o
1
A 2sin 40 cos 40
cos10
=
⇔
o
o
o o
1 cos10
A sin 80 1
cos10 cos10
= = =
Baøi 10 : Cho ABCΔ . Chöùng minh :
A B B C C A
tg tg tg tg tg tg 1
2 2 2 2 2 2
+ + =
Ta coù :
A B C
2 2
+ π
= −
2
Vaäy :
A B C
tg cot g
2 2
+
=
⇔
A B
tg tg
12 2
A B C
1 tg .tg tg
2 2 2
+
=
−
⇔
A B C A
tg tg tg 1 tg tg
2 2 2 2
⎡ ⎤
+ = −⎢ ⎥⎣ ⎦
B
2
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⇔
A C B C A B
tg tg tg tg tg tg 1
2 2 2 2 2 2
+ + =
Baøi 11 : Chöùng minh : ( )
π π π π
+ + + =8 4tg 2tg tg cot g *
8 16 32 32
Ta coù : (*) ⇔ 8 cot g tg 2tg 4tg
32 32 16 8
π π π
= − − −
π
Maø :
2 2
cosa sin a cos a sin a
cot ga tga
sin a cosa sin a cosa
−
− = − =
cos2a
2cot g2a
1
sin 2a
2
= =
Do ñoù :
(*) ⇔ cot g tg 2tg 4tg 8
32 32 16 8
π π π π⎡ ⎤
− − −⎢ ⎥⎣ ⎦
=
⇔ 2cot g 2tg 4tg 8
16 16 8
π π π⎡ ⎤
− −⎢ ⎥⎣ ⎦
=
⇔ 4cot g 4tg 8
8 8
π π
− =
⇔ 8cot g 8
4
π
= (hieån nhieân ñuùng)
Baøi :12 : Chöùng minh :
a/ 2 2 22 2
cos x cos x cos x
3 3
π π⎛ ⎞ ⎛ ⎞ 3
2
+ + + − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
b/
1 1 1 1
cot gx cot g16x
sin 2x sin 4x sin8x sin16x
+ + + = −
a/ Ta coù : 2 2 22 2
cos x cos x cos x
3 3
π π⎛ ⎞ ⎛
+ + + −⎜ ⎟ ⎜
⎝ ⎠ ⎝
⎞
⎟
⎠
( )
1 1 4 1 4
1 cos2x 1 cos 2x 1 cos 2x
2 2 3 2 3
⎡ π ⎤ ⎡ π ⎤⎛ ⎞ ⎛
= + + + + + + −⎜ ⎟ ⎜
⎞
⎟⎢ ⎥ ⎢
⎝ ⎠ ⎝
⎥
⎠⎣ ⎦ ⎣ ⎦
3 1 4 4
cos2x cos 2x cos 2x
2 2 3 3
⎡ π⎛ ⎞ ⎛
= + + + + −⎜ ⎟ ⎜⎢ ⎥
⎝ ⎠ ⎝⎣ ⎦
π ⎤⎞
⎟
⎠
3 1 4
cos2x 2cos2x cos
2 2 3
π⎡ ⎤
= + +⎢ ⎥⎣ ⎦
3 1 1
cos2x 2cos2x
2 2 2
⎡ ⎤⎛ ⎞
= + + −⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
3
2
=
b/ Ta coù :
cosa cos b sin bcosa sina cos b
cot ga cot gb
sina sin b sina sin b
−
− = − =
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( )sin b a
sin a sin b
−
=
Do ñoù :
( )
( )
sin 2x x 1
cot gx cot g2x 1
sin xsin 2x sin 2x
−
− = =
( )
( )
sin 4x 2x 1
cot g2x cot g4x 2
sin 2xsin 4x sin 4x
−
− = =
( )
( )
sin 8x 4x 1
cot g4x cot g8x 3
sin 4xsin 8x sin 8x
−
− = =
( )
( )
sin 16x 8x 1
cot g8x cot g16x 4
sin16xsin 8x sin16x
−
− = =
Laáy (1) + (2) + (3) + (4) ta ñöôïc
1 1 1 1
cot gx cot g16x
sin2x sin4x sin 8x sin16x
− = + + +
Baøi 13 : Chöùng minh : + =3 0 2 0
8sin 18 8sin 18 1
Ta coù: sin180
= cos720
⇔ sin180
= 2cos2
360
- 1
⇔ sin180
= 2(1 – 2sin2
180
)2
– 1
⇔ sin180
= 2(1 – 4sin2
180
+4sin4
180
)-1
⇔ 8sin4
180
– 8sin2
180
– sin180
+ 1 = 0 (1 )
⇔ (sin180
– 1)(8sin3
180
+ 8sin2
180
– 1) = 0
⇔ 8sin3
180
+ 8sin2
180
– 1 = 0 (do 0 < sin180
< 1)
Caùch khaùc :
Chia 2 veá cuûa (1) cho ( sin180
– 1 ) ta coù
( 1 ) ⇔ 8sin2
180
( sin180
+ 1 ) – 1 = 0
Baøi 14 : Chöùng minh :
a/ ( )4 4 1
sin x cos x 3 cos4x
4
+ = +
b/ ( )
1
sin6x cos6x 5 3cos4x
8
+ = +
c/ ( )8 8 1
sin x cos x 35 28cos4x cos8x
64
+ = + +
a/ Ta coù: ( )
24 4 2 2 2
sin x cos x sin x cos x 2sin x cos x+ = + − 2
22
1 sin 2
4
= − x
( )
1
1 1 cos4
4
= − − x
3 1
cos4x
4 4
= +
b/ Ta coù : sin6x + cos6x
( )( )2 2 4 2 2 4
sin x cos x sin x sin x cos x cos x= + − +
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( )4 4 21
sin x cos x sin 2x
4
= + −
(
3 1 1
cos4x 1 cos4x
4 4 8
⎛ ⎞
= + − −⎜ ⎟
⎝ ⎠
) ( do keát quaû caâu a )
3 5
cos4x
8 8
= +
c/ Ta coù : ( )+ = + −
28 8 4 4 4
sin x cos x sin x cos x 2sin x cos x4
( )= + −
2 41 2
3 cos4x sin 2x
16 16
( ) ( )⎡ ⎤
= + + − −⎢ ⎥⎣ ⎦
2
21 1 1
9 6cos4x cos 4x 1 cos4x
16 8 2
( ) ( )29 3 1 1
cos4x 1 cos8x 1 2cos4x cos 4x
16 8 32 32
= + + + − − +
( )= + + + − +
9 3 1 1 1
cos4x cos8x cos4x 1 cos8x
16 8 32 16 64
35 7 1
cos4x cos8x
64 16 64
= + +
Baøi 15 : Chöùng minh : 3 3
sin3x.sin x cos3x.cos x cos 2x+ = 3
Caùch 1:
Ta coù : s 3 3 3
in3x.sin x cos3x.cos x cos 2x+ =
( ) ( )3 3 3 3
3sin x 4sin x sin x 4 cos x 3cos x cos x= − + −
4 6 6
3sin x 4sin x 4cos x 3cos x= − + − 4
( ) ( )4 4 6 6
3 sin x cos x 4 sin x cos x= − − −
( )( )2 2 2 2
3 sin x cos x sin x cos x= − +
( )( )2 2 4 2 2 4
4 sin x cos x sin x sin x cos x cos x− − + +
2 2
3cos2x 4 cos2x 1 sin x cos x⎡ ⎤= − + −⎣ ⎦
21
3cos2x 4 cos2x 1 sin 2x
4
⎛ ⎞
= − + −⎜ ⎟
⎝ ⎠
21
cos2x 3 4 1 sin 2x
4
⎡ ⎤⎛ ⎞
= − + −⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
( )2
cos2x 1 sin 2x= −
3
cos 2x=
Caùch 2 :
Ta coù : si 3 3
n3x.sin x cos3x.cos x+
3sin x sin 3x 3cos x cos3x
sin 3x cos3x
4 4
− +⎛ ⎞ ⎛
= +⎜ ⎟ ⎜
⎝ ⎠ ⎝
⎞
⎟
⎠
( ) ( )2 23 1
sin3xsin x cos3xcos x cos 3x sin 3x
4 4
= + + −
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( )
3 1
cos 3x x cos6x
4 4
= − +
( )
1
3cos2x cos3.2x
4
= +
( )= + −31
3cos2x 4cos 2x 3cos2x
4
( boû doøng naøy cuõng ñöôïc)
3
cos 2x=
Baøi 16 : Chöùng minh : o o o o o 3 1
cos12 cos18 4 cos15 .cos21 cos24
2
+
+ − = −
Ta coù : ( )o o o o
cos12 cos18 4 cos15 cos21 cos24+ − o
( )o o o o
2cos15 cos3 2cos15 cos45 cos3= − + o
o o o o o o
2cos15 cos3 2cos15 cos45 2cos15 cos3= − −
o o
2cos15 cos45= −
( )o o
cos60 cos30= − +
3 1
2
+
= −
Baøi 17 : Tính 2 o 2 o
P sin 50 sin 70 cos50 cos70= + − o
Ta coù : ( ) ( ) ( )= − + − − +o o o1 1 1
P 1 cos100 1 cos140 cos120 cos20
2 2 2
o
( )o o1 1 1
P 1 cos100 cos140 cos20
2 2 2
⎛ ⎞
= − + − − +⎜ ⎟
⎝ ⎠
o
( )o o 1 1
P 1 cos120 cos20 cos20
4 2
= − + − o
o o5 1 1 5
P cos20 cos20
4 2 2 4
= + − =
Baøi 18 : Chöùng minh : o o o o 8 3
tg30 tg40 tg50 tg60 cos20
3
+ + + = o
AÙp duïng :
( )sin a b
tga tgb
cosa cos b
+
+ =
Ta coù : ( ) ( )o o o o
tg50 tg40 tg30 tg60+ + +
o o
o o o
sin 90 sin 90
cos50 cos40 cos30 cos60
= + o
o o
o
1 1
1sin 40 cos40 cos30
2
= +
o o
2 2
sin80 cos30
= +
o o
1 1
2
cos10 cos30
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
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o o
o o
cos30 cos10
2
cos10 cos30
⎛ ⎞+
= ⎜ ⎟
⎝ ⎠
p o
o o
cos20 cos10
4
cos10 cos30
=
o8 3
cos20
3
=
Baøi 19 : Cho ABCΔ , Chöùng minh :
a/
A B C
sin A sinB sinC 4cos cos cos
2 2
+ + =
2
b/
A B C
socA cosB cosC 1 4sin sin sin
2 2
+ + = +
2
c/ sin 2A sin 2B sin 2C 4 sin A sin Bsin C+ + =
d/ 2 2 2
cos A cos B cos C 2cos A cosBcosC+ + = −
e/ tgA tgB tgC tgA.tgB.tgC+ + =
f/ cot gA.cot gB cot gB.cot gC cot gC.cot gA 1+ + =
g/ + + =
A B C A B
cot g cot g cot g cot g .cot g .cot g
2 2 2 2 2
C
2
a/ Ta coù : ( )
A B A B
sin A sinB sinC 2sin cos sin A B
2 2
+ −
+ + = + +
A B A B A B
2sin cos cos
2 2 2
+ − +⎛ ⎞
= +⎜ ⎟
⎝ ⎠
+ π⎛ ⎞
= =⎜ ⎟
⎝ ⎠
C A B A B C
4 cos cos cos do
2 2 2 2 2 2
−
b/ Ta coù : ( )
A B A B
cos A cosB cosC 2cos cos cos A B
2 2
+ −
+ + = − +
2A B A B A B
2cos cos 2cos 1
2 2 2
+ − +⎛ ⎞
= − ⎜ ⎟
⎝ ⎠
−
A B A B A B
2cos cos cos 1
2 2 2
+ − +⎡ ⎤
= −⎢ ⎥⎣ ⎦
+
A B A B
4 cos sin sin 1
2 2 2
+ ⎛ ⎞
= − − +⎜ ⎟
⎝ ⎠
C A B
4sin sin sin 1
2 2 2
= +
c/ ( ) ( )sin 2A sin 2B sin 2C 2sin A B cos A B 2sinCcosC+ = + − +
= − +2sinCcos(A B) 2sinCcosC
= − −2sinC[cos(A B) cos(A B)]+
2
= − −4sinCsin A sin( B)
= 4 sin Csin A sin B
d/ + +2 2
cos A cos B cos C
( ) 21
1 cos2A cos2B cos C
2
= + + +
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( ) ( ) 2
1 cos A B cos A B cos C= + + − +
( )1 cosC cos A B cosC= − − −⎡ ⎤⎣ ⎦ do ( )( )cos A B cosC+ = −
( ) ( )1 cosC cos A B cos A B= − − + +⎡ ⎤⎣ ⎦
1 2cosC.cos A.cosB= −
e/ Do neân ta coùa b C+ = π −
( )tg A B tgC+ = −
⇔
tgA tgB
tgC
1 tgAtgB
+
= −
−
⇔ tgA tgB tgC tgAtgBtgC+ = − +
⇔ tgA tgB tgC tgAtgBtgC+ + =
f/ Ta coù : cotg(A+B) = - cotgC
⇔
1 tgAtgB
cot gC
tgA tgB
−
= −
+
⇔
cot gA cot gB 1
cot gC
cot gB cot gA
−
= −
+
(nhaân töû vaø maãu cho cotgA.cotgB)
⇔ cot gA cot gB 1 cot gCcot gB cot gA cot gC− = − −
⇔ cot gA cot gB cot gBcot gC cot gA cot gC 1+ + =
g/ Ta coù :
A B C
tg cot g
2 2
+
=
⇔
A B
tg tg
C2 2 cot g
A B 21 tg tg
2 2
+
=
−
⇔
A B
cot g cot g
C2 2 cot g
A B 2cot g .cot g 1
2 2
+
=
−
(nhaân töû vaø maãu cho cotg
A
2
.cotg
B
2
)
⇔
A B A B C
cot g cot g cot g cot g cot g cot g
2 2 2 2 2
+ = −
C
2
⇔
A B C A B
2 2 2 2 2
+ + =
C
2
cos2A + cos2B + cos 2C + 4cosAcosBcosC + 1 = 0
Ta coù : (cos2A + cos2B) + (cos2C + 1)
= 2 cos (A + B)cos(A - B) + 2cos2
C
= - 2cosCcos(A - B) + 2cos2
C
= - 2cosC[cos(A – B) + cos(A + B)] = - 4cosAcosBcosC
Do ñoù : cos2A + cos2B + cos2C + 1 + 4cosAcosBcosC = 0
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cos3A + cos3B + cos3C = 1 -
3A 3B 3C
4sin sin sin
2 2 2
Ta coù : (cos3A + cos3B) + cos3C
23 3
2cos (A B)cos (A B) 1 2sin
2 2
= + − + −
3C
2
Maø : A B C+ = π − neân ( )
3 3
A B
2 2
+ = π −
3C
2
=> ( )
3 3
cos A B cos
2 2
π⎛ ⎞
+ = −⎜ ⎟
⎝ ⎠
3C
2
3C
cos
2 2
π⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
3C
sin
2
= −
Do ñoù : cos3A + cos3B + cos3C
( ) 23 A B3C 3C
2sin cos 2sin 1
2 2 2
−
= − − +
( )3 A B3C 3C
2sin cos sin 1
2 2 2
−⎡ ⎤
= − + +⎢ ⎥
⎣ ⎦
( )
( )
3 A B3C 3
2sin cos cos A B 1
2 2 2
−⎡ ⎤
= − − + +⎢ ⎥
⎣ ⎦
−
= +
3C 3A 3B
4sin sin sin( ) 1
2 2 2
3C 3A 3B
4sin sin sin 1
2 2 2
= − +
Baøi 22 : A, B, C laø ba goùc cuûa moät tam giaùc. Chöùng minh :
sin A sin B sinC A B C
tg tg cot g
cos A cosB cosC 1 2 2 2
+ −
=
+ − +
Ta coù :
2
A B A B C C
2sin cos 2sin cos
sin A sin B sinC 2 2 2
A B A B Ccos A cosB cosC 1 2cos cos 2sin
2 2 2
2
+ −
−
+ −
=
+ −+ − + +
C A B C A B A2cos cos sin cos cos
C2 2 2 2 2cot g .
B
A B AC A B C 2 cos cos2sin cos sin
2 22 2 2
−⎡ ⎤
B
− +− −⎢ ⎥⎣ ⎦= =
− +−⎡ ⎤ ++⎢ ⎥⎣ ⎦
A B
2sin .sin
C 2 2
cot g .
A B2 2cos .cos
2 2
⎛ ⎞
− −⎜ ⎟
⎝ ⎠=
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C A B
cot g .tg .tg
2 2 2
=
A B C B C A C A B
sin cos cos sin cos cos sin cos cos
2 2 2 2 2 2 2 2 2
+ +
( )
A B C A B B C A C
sin sin sin tg tg tg tg tg tg *
2 2 2 2 2 2 2 2 2
= + + +
Ta coù :
A B C
2 2 2
+ π
= − vaäy
A B C
tg cot g
2 2 2
⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
⇔
A B
tg tg
12 2
A B C
1 tg tg tg
2 2 2
+
=
−
⇔
A B C A
tg tg tg 1 tg tg
2 2 2 2
⎡ ⎤
+ = −⎢ ⎥⎣ ⎦
B
2
⇔ ( )
A C B C A B
tg tg tg tg tg tg 1 1
2 2 2 2 2 2
+ + =
Do ñoù : (*)
A B C B C A C A
2 2 2 2 2 2 2 2 2
+ +
B
A B C
sin sin sin 1
2 2 2
= + (do (1))
⇔
A B C B C A B C C B
sin cos cos sin sin cos sin cos sin cos 1
2 2 2 2 2 2 2 2 2 2
⎡ ⎤ ⎡
− + +⎢ ⎥ ⎢⎣ ⎦ ⎣
⎤
=⎥⎦
⇔
A B C A B C
sin cos cos sin 1
2 2 2 2
+ +
+ =
⇔
A B C
sin 1
2
+ +
=
π
⇔ sin 1
2
= ( hieån nhieân ñuùng)
Baøi 24 : Chöùng minh : ( )
A B C 3 cos A cosB cosC
tg tg tg *
2 2 2 sin A sinB sinC
+ + +
+ + =
+ +
Ta coù :
2A B A B C
cos A cosB cosC 3 2cos cos 1 2sin 3
2 2 2
+ − ⎡ ⎤
+ + + = + − +⎢ ⎥⎣ ⎦
2C A B
2sin cos 4 2sin
2 2
C
2
−
= + −
C A B C
2sin cos sin 4
2 2 2
−⎡ ⎤
= − +⎢ ⎥⎣ ⎦
C A B A B
2sin cos cos 4
2 2 2
− +⎡ ⎤
= − +⎢ ⎥⎣ ⎦
C A B
4sin sin .sin 4
2 2 2
= + (1)
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A B A B
sin A sinB sinC 2sin cos sinC
2 2
+ −
+ + = +
C A B C
2cos cos 2sin cos
2 2 2
C
2
−
= +
C A B A B
2cos cos cos
2 2 2
− +⎡ ⎤
= +⎢ ⎥⎣ ⎦
C A
4cos cos cos
2 2
=
B
2
(2)
Töø (1) vaø (2) ta coù :
(*) ⇔
A B C A B C
sin sin sin sin sin sin 1
2 2 2 2 2 2
A B C A B C
cos cos cos cos cos cos
2 2 2 2 2 2
+
+ + =
⇔
A B C B A C C A B
2 2 2 2 2 2 2 2 2
⎡ ⎤ ⎡ ⎤ ⎡
+ +⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣
⎤
⎥⎦
A B C
sin sin sin 1
2 2 2
= +
⇔
A B C B C A B C C B
sin cos cos sin sin cos sin cos sin cos 1
2 2 2 2 2 2 2 2 2 2
⎡ ⎤ ⎡
− + +⎢ ⎥ ⎢⎣ ⎦ ⎣
⎤
=⎥⎦
⇔
A B C A B C
sin .cos cos sin 1
2 2 2 2
+ +
+ =
⇔
A B C
sin 1
2
+ +⎡ ⎤
=⎢ ⎥⎣ ⎦
⇔ sin 1
2
π
= ( hieån nhieân ñuùng)
Baøi 25 : Cho ABCΔ . Chöùng minh:
A B C
sin sin sin
2 2 2 2
B C C A A B
2 2 2 2 2 2
+ + =
Caùch 1 :
A B A A B B
sin sin sin cos sin cos
2 2 2 2 2 2
B C C A A B C
cos cos cos cos cos cos cos
2 2 2 2 2 2 2
+
+ =Ta coù :
A B A
sin cos
1 sin A sin B 2 2
B
A B C A B C2 cos cos cos cos cos cos
2 2 2 2 2 2
+ −
+
= =
−⎛ ⎞−
⎜ ⎟
⎝ ⎠= =
A BC A B coscos .cos
22 2
A B C A
cos .cos .cos cos cos
2 2 2 2
B
2
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Do ñoù : Veá traùi
A B C A B Acos sin cos cos
2 2 2
A B A B A B
2 2 2 2 2 2
−⎛ ⎞ B
2
− +
+⎜ ⎟
⎝ ⎠= + =
A B
2cos cos
2 2 2
A B
cos cos
2 2
= =
Caùch 2 :
Ta coù veá traùi
B C A C A B
cos cos cos
2 2
B C C A A B
2 2 2 2 2
+ + +
= + + 2
2
B C B C A C A C
cos cos sin sin cos cos sin sin
2 2 2 2 2 2 2
B C C A
cos cos cos cos
2 2 2 2
− −
= + 2
A B A
cos cos sin sin
2 2 2
A B
cos cos
2 2
−
+
B
2
B C A C A B
3 tg tg tg tg tg tg
2 2 2 2 2 2
⎡ ⎤
= − + +⎢ ⎥⎣ ⎦
Maø :
A B B C A B
tg tg tg tg tg tg 1
2 2 2 2 2 2
+ + =
(ñaõ chöùng minh taïi baøi 10 )
Do ñoù : Veá traùi = 3 – 1 = 2
Baøi 26 : Cho ABCΔ . Coù
A B
cot g ,cot g ,cot g
2 2
C
2
theo töù töï taïo caáp soá coäng.
Chöùng minh
A C
cot g .cot g 3
2 2
=
Ta coù :
A B
cot g ,cot g ,cot g
2 2
C
2
laø caáp soá coäng
⇔
A C B
cot g cot g 2cot g
2 2
+ =
2
⇔
+
=
A C B
sin 2cos
2 2
A C B
sin sin sin
2 2 2
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⇔
B B
cos 2cos
2 2
A C B
sin sin sin
2 2 2
=
⇔ =
+
1 2
A C A
sin sin cos
2 2 2
C
(do 0<B<π neân
B
cos 0
2
> )
⇔
A C A C
cos cos sin sin
2 2 2 2 2
A C
sin .sin
2 2
−
= ⇔
A C
cot g cot g 3
2 2
=
1 1 1 1 A B C A B C
tg tg tg cot g cot g cot g
sin A sin B sinC 2 2 2 2 2 2 2
⎡ ⎤
+ + = + + + + +⎢ ⎥⎣ ⎦
Ta coù :
A B C A B
2 2 2 2 2
+ + =
C
2
(Xem chöùng minh baøi 19g )
Maët khaùc :
sin cos 2
tg cot g
cos sin sin2
α α
α + α = + =
α α α
Do ñoù :
1 A B C A B C
tg tg tg cotg cotg cotg
2 2 2 2 2 2 2
⎡ ⎤
+ + + + +⎢ ⎥⎣ ⎦
1 A B C 1 A B C
tg tg tg cotg cotg cotg
2 2 2 2 2 2 2 2
⎡ ⎤ ⎡
= + + + + +
⎤
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
1 A A 1 B B 1 C C
tg cot g tg cot g tg cot g
2 2 2 2 2 2 2 2 2
⎡ ⎤ ⎡ ⎤ ⎡
= + + + + +
⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
1 1 1
sin A sinB sinC
= + +
BAØI TAÄP
1. Chöùng minh :
a/
2 1
cos cos
5 5
π π
− =
2
b/
o o
o o
cos15 sin15
3
cos15 sin15
+
=
−
c/
2 4 6
cos cos cos
7 7 7
π π π
+ + =
1
2
−
3
d/ + =3 3
sin 2xsin6x cos 2x.cos6x cos 4x
e/ o o o o
tg20 .tg40 .tg60 .tg80 3=
f/
π π π π
+ + + =
2 5 8 3
tg tg tg tg cos
6 9 18 3 3
π
9
g/ 7
2 3 4 5 6 7
cos .cos .cos .cos .cos .cos .cos
15 15 15 15 15 15 15 2
π π π π π π π
=
1
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h/ tgx.tg x .tg x tg3x
3 3
π π⎡ ⎤ ⎡ ⎤
− + =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
k/ o o o o
tg20 tg40 3tg20 .tg40 3+ + =
e/ o o o 3
sin 20 .sin 40 .sin 80
8
=
m/ o o o o
tg5 .tg55 .tg65 .tg75 1=
2. Chöùng minh raèng neáu
( )
( ) ( )
sin x 2sin x y
x y 2k 1 k z
2
= +⎧
⎪
⎨ π
+ ≠ + ∈⎪
⎩
thì
sin
( )
cos
y
tg x y
y
+ =
− 2
3. Cho ABCΔ coù 3 goùc ñeàu nhoïn vaø A B C≥ ≥
a/ Chöùng minh : tgA + tgB + tgC = tgA.tgB.tgC
b/ Ñaët tgA.tgB = p; tgA.tgC = q
Chöùng minh (p-1)(q-1)≥ 4
4. Chöùng minh caùc bieåu thöùc khoâng phuï thuoäc x :
a/ ( ) ( )4 2 4 2 2 2
A sin x 1 sin x cos x 1 cos x 5sin x cos x 1= + + + + +
b/ ( ) ( )8 8 6 6
B 3 sin x cos x 4 cos x 2sin x 6sin x= − + − + 4
c/ ( ) ( ) ( ) ( ) (2 2
C cos x a sin x b 2cos x a sin x b sin a b= − + − − − − − )
5. Cho ABCΔ , chöùng minh :
a/
cosC cosB
cot gB cot gC
sinBcos A sinCcos A
+ = +
b/ 3 3 3 A B C 3A 3B 3
sin A sin B sin C 3cos cos cos cos cos cos
2 2 2 2 2 2
+ + = +
C
c/
A B C B A C
sin A sinB sinC cos .cos cos .cos
2 2 2 2
− −
+ + = +
C A
cos .cos
2 2
B−
+
d/ cotgAcotgB + cotgBcotgC + cotgCcotgA = 1
e/ 2 2 2
cos A cos B cos C 1 2cos A cosBcosC+ + = −
f/ sin3Asin(B- C)+ sin3Bsin(C- A)+ sin3Csin(A- B) = 0
6. Tìm giaù trò nhoû nhaát cuûa :
a/
1 1
y
sin x cos x
= + vôùi 0 x
2
π
< <
b/
π
= + +
9
y 4x sin x
x
vôùi 0 x< < ∞
c/ 2
y 2sin x 4sin x cos x 5= + +
7. Tìm giaù trò lôùn nhaát cuûa :
a/ y sin x cos x cos x sin x= +
b/ y = sinx + 3sin2x
c/ 2
y cos x 2 cos x= + −
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Chöông 2 : PHÖÔNG TRÌNH LÖÔÏNG GIAÙC CÔ BAÛN
= + π⎡
= ⇔ ⎢ = π − + π⎣
u v k2
sin u sin v
u v k2
cosu cos v u v k2= ⇔ = ± + π
π⎧
≠ + π⎪
= ⇔ ⎨
⎪ = + π⎩
u k
tgu tgv 2
u v k '
( )k,k ' Z∈
u k
cot gu cot gv
u v k '
≠ π⎧
= ⇔ ⎨
= + π⎩
Ñaëc bieät : sinu 0 u k= ⇔ = π
π
= ⇔ = + πcosu 0 u k
2
(sinu 1 u k2 k Z
2
π
= ⇔ = + π ∈ ) cosu 1 u k2= ⇔ = π ( )k Z∈
sin u 1 u k2
2
π
= − ⇔ = − + π cosu 1 u k2= − ⇔ = π + π
Chuù yù : sinu 0 cosu 1≠ ⇔ ≠ ±
cosu 0 sinu 1≠ ⇔ ≠ ±
Baøi 28 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái D, naêm 2002)
[ ]x 0,14∈ nghieäm ñuùng phöông trìnhTìm
( )cos3x 4 cos2x 3cos x 4 0 *− + − =
Ta coù (*) : ⇔ ( ) ( )3 2
4 cos x 3cos x 4 2cos x 1 3cos x 4 0− − − + − =
⇔ 3 2
4cos x 8cos x 0− = ⇔ ( )2
4 cos x cos x 2 0− =
⇔ ( )= =cos x 0 hay cos x 2 loaïi vìcos x 1≤
⇔ ( )x k k
2
π
= + π ∈ Z
Ta coù : [ ]x 0,14 0 k 1
2
4
π
∈ ⇔ ≤ + π ≤
⇔ k 14
2 2
π π
− ≤ π ≤ − ⇔
1 14 1
0,5 k 3,9
2 2
− = − ≤ ≤ − ≈
π
Maø neânk Z∈ { }k 0,1,2,3∈ . Do ñoù :
3 5 7
x , , ,
2 2 2 2
π π π π⎧ ⎫
∈ ⎨ ⎬
⎩ ⎭
Giaûi phöông trình :
( )( ) ( )2cos x 1 2sin x cos x sin 2x sin x *− + = −
Ta coù (*) ⇔ ( )( ) ( )− + =2cos x 1 2sin x cos x sin x 2cos x 1−
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⇔ ( ) ( )2cos x 1 2sin x cos x sin x 0− + −⎡ ⎤⎣ ⎦ =
)⇔ ( )(2cos x 1 sin x cos x 0− + =
⇔
1
cos x sin x cos x
2
= ∨ = −
⇔ cos x cos tgx 1 tg
3 4
π π⎛ ⎞
= ∨ = − = −⎜ ⎟
⎝ ⎠
⇔ ( )
π π
= ± + π ∨ = − + π ∈x k2 x k , k
3 4
Z
Baøi 30 : Giaûi phöông trình + + + =cos x cos2x cos3x cos4x 0(*)
Ta coù (*) ⇔ ( ) ( )cos x cos4x cos2x cos3x 0+ + + =
⇔
5x 3x 5x x
2cos .cos 2cos .cos 0
2 2 2 2
+ =
⇔
5x 3x x
2cos cos cos 0
2 2 2
⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
⇔
5x x
4cos cos x cos 0
2 2
=
⇔
5x x
cos 0 cos x 0 cos 0
2 2
= ∨ = ∨ =
⇔
π π π
= + π ∨ = + π ∨ = + π
5x x
k x k k
2 2 2 2 2
⇔ ( )
π π π
= + ∨ = + π ∨ = π + π ∈
2k
x x k x 2 ,
5 5 2
k Z
Baøi 31: Giaûi phöông trình ( )2 2 2 2
sin x sin 3x cos 2x cos 4x *+ = +
Ta coù (*) ⇔ ( ) ( ) ( ) ( )
1 1 1 1
1 cos2x 1 cos6x 1 cos4x 1 cos8x
2 2 2 2
− + − = + + +
⇔ ( )cos2x cos6x cos4x cos8x− + = +
⇔ 2cos4xcos2x 2cos6xcos2x− =
⇔ ( )2cos2x cos6x cos4x 0+ =
⇔ 4cos2xcos5xcos x 0=
⇔ cos2x 0 cos5x 0 cos x 0= ∨ = ∨ =
⇔
π π π
= + π ∨ + π ∨ = + π ∈2x k 5x k x k ,k
2 2 2
⇔
π π π π π
= + ∨ = + ∨ = + π
k k
x x x k
4 2 10 5 2
∈, k
Baøi 32 : Cho phöông trình
( )
π⎛ ⎞
− = − −⎜ ⎟
⎝ ⎠
2 2 x 7
sin x.cos4x sin 2x 4sin *
4 2 2
Tìm caùc nghieäm cuûa phöông trình thoûa: − <x 1 3
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Ta coù : (*)⇔ ( )
1 7
sin x.cos4x 1 cos4x 2 1 cos x
2 2
⎡ π ⎤⎛ ⎞
2
− − = − −⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
−
⇔ − + = − −
1 1 3
sin x cos4x cos4x 2sin x
2 2 2
⇔
1
sin xcos4x cos4x 1 2sin x 0
2
+ + + =
⇔
⎛ ⎞ ⎛ ⎞
+ + + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1 1
cos4x sin x 2 sin x 0
2 2
⇔ ( )
1
cos4x 2 sin x 0
2
⎛ ⎞
+ + =⎜ ⎟
⎝ ⎠
⇔
( )cos4x 2 loaïi
1
sin x sin
2 6
= −⎡
⎢
π⎛⎢ = − = −⎜ ⎟⎢ ⎝ ⎠⎣
⎞ ⇔
π⎡
= − + π⎢
⎢
π⎢ = + π
⎢⎣
x k
6
7
x 2
6
2
h
Ta coù : − <x 1 3 ⇔ ⇔3 x 1 3− < − < 2 x 4− < <
Vaäy : 2 k2
6
π
− < − + π < 4
⇔ 2 2k 4
6 6
π π
− < π < + ⇔
1 1 2 1
k
12 12
− < < +
π π
Do k neân k = 0. VaäyZ∈ x
6
π
= −
π
− < + π <
7
2 h2
6
4
⇔
π π
− − < π < − ⇔ − − < < −
π π
7 7 1 7 2
2 h2 4 h
6 6 12
7
12
⇒ h = 0 ⇒
π
=
7
x
6
.Toùm laïi
−π π
= =
7
x hay x
6 6
Caùch khaùc :
−π
= − ⇔ = − + π ∈k1
sin x x ( 1) k , k
2 6
Vaäy :
−π − −
− < − + π < ⇔ < − + <
π π
k k2 1
2 ( 1) k 4 ( 1) k
6 6
4
⇔ k=0 vaø k = 1. Töông öùng vôùi
−π π
= =
7
x hay x
6 6
Baøi 33 : Giaûi phöông trình
( )3 3 3
sin x cos3x cos x sin 3x sin 4x *+ =
Ta coù : (*)⇔ ( ) ( )3 3 3 3 3
sin x 4 cos x 3cos x cos x 3sin x 4sin x sin 4x− + − =
⇔ 3 3 3 3 3 3 3
4sin xcos x 3sin xcos x 3sin xcos x 4sin xcos x sin 4x− + − =
⇔ ( )2 2 3
3sin x cos x cos x sin x sin 4x− =
⇔ 33
sin2x cos2x sin 4x
2
=
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⇔ 33
sin 4x sin 4x
4
=
⇔ 3
3sin4x 4sin 4x 0− =
⇔ sin12x = 0
⇔ ⇔12x k= π ( )
k
x k
12
Z
π
= ∈
Baøi 34 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái B, naêm 2002)
( )2 2 2 2
sin 3x cos 4x sin 5x cos 6a *− = −
Ta coù : (*)⇔
( ) ( ) ( ) ( )
1 1 1 1
2 2 2 2
− − + = − − +
⇔ cos6x cos8x cos10x cos12x+ = +
⇔ 2cos7xcos x 2cos11xcos x=
⇔ ( )2cos x cos7x cos11x 0− =
⇔ cos x 0 cos7x cos11x= ∨ =
⇔
π
= + π ∨ = ± + πx k 7x 11x k
2
2
⇔
π π π
= + π ∨ = − ∨ = ∈
k k
x k x x , k
2 2 9
( ) ( )sin x sin 3x sin 2x cos x cos3x cos2x+ + = + +
⇔ 2sin2xcos x sin2x 2cos2xcos x cos2x+ = +
⇔ ( ) ( )+ = +sin 2x 2 cos x 1 cos 2x 2 cos x 1
⇔ ( )( )2cos x 1 sin 2x cos2x 0+ − =
⇔
1 2
cos x cos sin 2x cos2x
2 3
π
= − = ∨ =
⇔
2
x k2 tg2x 1
3 4
tg
π π
= ± + π ∨ = =
⇔ ( )
π π π
= ± + π ∨ = + ∈
2
x k2 x k , k
3 8 2
Z
Baøi 36: Giaûi phöông trình
( )+ + = +2 3
cos10x 2 cos 4x 6 cos 3x.cos x cos x 8 cos x.cos 3x *
Ta coù : (*)⇔ ( ) ( )3
cos10x 1 cos8x cos x 2cos x 4 cos 3x 3cos3x+ + = + −
⇔ ( )cos10x cos8x 1 cos x 2cos x.cos9x+ + = +
⇔ 2cos9xcos x 1 cos x 2cos x.cos9x+ = +
⇔ cos x 1=
⇔ ( )x k2 k Z= π ∈
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( )3 3 2
4 sin x 3cos x 3sin x sin x cos x 0 *+ − − =
Ta coù : (*) ⇔ ( ) ( )2 2
sin x 4sin x 3 cos x sin x 3cos x 02
− − − =
⇔ ( ) ( )⎡ ⎤− − − − =⎣ ⎦
2 2
sin x 4 sin x 3 cos x sin x 3 1 sin x 02
=
=
⇔ ( )( )2
4sin x 3 sin x cos x 0− −
⇔ ( ) ( )2 1 cos2x 3 sin x cos x 0− − −⎡ ⎤⎣ ⎦
⇔
1 2
cos2x cos
2 3
sin x cos x
π⎡
= − =⎢
⎢
=⎣
⇔
2
2x k2
3
tgx 1
π⎡
= ± + π⎢
⎢
=⎣
⇔
x k
3
x k
4
π⎡
= ± + π⎢
⎢
π⎢ = + π
⎢⎣
( )k Z∈
Baøi 38 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái B naêm 2005)
( )sin x cos x 1 sin 2x cos2x 0 *+ + + + =
Ta coù : (*) ⇔ 2
sin x cos x 2sin xcos x 2cos x 0+ + + =
⇔ ( )sin x cos x 2cos x sin x cos x 0+ + + =
⇔ ( )(sin x cos x 1 2cos x 0+ + ) =
⇔
sin x cos x
1 2
cos2x cos
2 3
= −⎡
⎢ π⎢ = − =
⎣
⇔
tgx 1
2
x k
3
= −⎡
⎢ π⎢ = ± + π
⎣
2
⇔
x k
4
2
x k2
3
π⎡
= − + π⎢
⎢
π⎢ = ± + π
⎢⎣
k Z∈( )
( )( ) ( )2
2sin x 1 3cos4x 2sin x 4 4 cos x 3 *+ + − + =
Ta coù : (*) ⇔ ( )( ) ( )2
2sin x 1 3cos4x 2sin x 4 4 1 sin x 3 0+ + − + − − =
⇔ ( )( ) ( )( )2sin x 1 3cos4x 2sin x 4 1 2sin x 1 2sin x 0+ + − + + − =
⇔ ( ) ( )2sin x 1 3cos4x 2sin x 4 1 2sin x 0+ + − + −⎡ ⎤⎣ ⎦ =
=⇔ ( )( )3 cos4x 1 2sin x 1 0− +
⇔
1
cos4x 1 sin x sin
2 6
π⎛ ⎞
= ∨ = − = −⎜ ⎟
⎝ ⎠
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⇔
π π
= π ∨ = − + π ∨ = +
7
4x k2 x k2 x k2
6 6
π
⇔ ( )
π π π
= ∨ = − + π ∨ = + π ∈
k 7
x x k2 x k2 , k
2 6 6
Z
( )( )+ = +6 6 8 8
sin x cos x 2 sin x cos x *
Ta coù : (*) ⇔
6 8 6 8
sin x 2sin x cos x 2cos x 0− + − =
⇔ ( ) ( )6 2 6 2
sin x 1 2sin x cos x 2cos x 1 0− − − =
⇔ − =6 6
sin x cos 2x cos x.cos 2x 0
⇔ ( )6 6
cos2x sin x cos x 0− =
⇔ 6 6
cos2x 0 sin x cos x= ∨ =
⇔ 6
cos2x 0 tg x 1= ∨ =
⇔ ( )2x 2k 1 tgx 1
2
π
= + ∨ = ±
⇔ ( )x 2k 1 x k
4 4
π π
= + ∨ = ± + π
⇔
k
x
4 2
π π
= + ,k ∈
( )
1
cos x.cos2x.cos4x.cos8x *
16
=
Ta thaáy x k= π khoâng laø nghieäm cuûa (*) vì luùc ñoù
cos x 1,cos2x cos4x cos8x 1= ± = = =
(*) thaønh :
1
1
16
± = voâ nghieäm
Nhaân 2 veá cuûa (*) cho 16sin x 0≠ ta ñöôïc
(*)⇔ vaø si( )16sin x cos x cos2x.cos4x.cos8x sin x= n x 0≠
⇔ vaø si( )8sin 2x cos2x cos4x.cos8x sin x= n x 0≠
⇔ vaø si( )4sin 4x cos4x cos8x sin x= n x 0≠
⇔ vaø2sin8xcos8x sin x= sin x 0≠
⇔ sin16x sin x= vaø sin x 0≠
⇔ ( )
π π π
= ∨ = + ∈
k2 k
x x , k
15 17 17
Z
Do : khoâng laø nghieäm neân= πx h ≠k 15m vaø ( )+ ≠ ∈2k 1 17n n, m Z
Baøi 42: Giaûi phöông trình ( )3
8cos x cos 3x *
3
π
+ =
⎛ ⎞
⎜ ⎟
⎝ ⎠
Ñaët t x x t
3 3
π π
= + ⇔ = −
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Thì ( ) ( )cos 3x cos 3t cos 3t cos 3t= − π = π − = −
Vaäy (*) thaønh = −3
8cos t cos3t
⇔ 3 3
8cos t 4 cos t 3cos t= − +
⇔ 3
12cos t 3cos t 0− =
⇔ ( )2
3cos t 4 cos t 1 0− =
⇔ ( )3cos t 2 1 cos 2t 1 0+ −⎡ ⎤⎣ ⎦ =
⇔ ( )cos t 2 cos 2t 1 0+ =
⇔
1 2
cos t 0 cos 2t cos
2 3
π
= ∨ = − =
⇔ ( )
π π
= + ∨ = ± +
2
t 2k 1 2t k2
2 3
π
⇔
π π
= + π ∨ = ± + πt k t
2 3
k
Maø x t
3
π
= −
Vaäy (*)⇔ ( )
π π
= + π ∨ = π ∨ = + π ∈
2
x k2 x k x k , vôùik
6 3
Z
Ghi chuù :
Khi giaûi caùc phöông trình löôïng giaùc coù chöùa tgu, cotgu, coù aån ôû maãu, hay
chöùa caên baäc chaün... ta phaûi ñaët ñieàu kieän ñeå phöông trình xaùc ñònh. Ta seõ
duøng caùc caùch sau ñaây ñeå kieåm tra ñieàu kieän xem coù nhaän nghieäm hay
khoâng.
+ Thay caùc giaù trò x tìm ñöôïc vaøo ñieàu kieän thöû laïi xem coù thoûa
Hoaëc + Bieåu dieãn caùc ngoïn cung ñieàu kieän vaø caùc ngoïn cung tìm ñöôïc treân cuøng
moät ñöôøng troøn löôïng giaùc. Ta seõ loaïi boû ngoïn cung cuûa nghieäm khi coù
truøng vôùi ngoïn cung cuûa ñieàu kieän.
Hoaëc + So vôi caùc ñieàu kieän trong quaù trình giaûi phöông trình.
Baøi 43 : Giaûi phöông trình ( )2
tg x tgx.tg3x 2 *− =
Ñieàu kieän 3
cos x 0
cos 3x 4 cos x 3cos x 0
≠⎧
⎨
= − ≠⎩
π π
⇔ ≠ ⇔ ≠ +
h
cos3x 0 x
6 3
Luùc ñoù ta coù (*) ⇔ ( )tgx tgx tg3x 2− =
⇔
sin x sin x sin 3x
2
cos x cos x cos 3x
⎛ ⎞
− =⎜ ⎟
⎝ ⎠
⇔ ( ) 2
sin x sin x cos 3x cos x sin 3x 2 cos x cos 3x− =
⇔ ( ) 2
sin x sin 2x 2 cos x.cos 3x− =
⇔ 2 2
2sin x cos x 2cos x cos 3x− =
⇔ (do cos2
sin x cos x cos 3x− = x 0≠ )
⇔ ( ) ( )
1 1
1 cos 2x cos 4x cos 2x
2 2
− − = +
⇔ cos 4x 1 4x k2= − ⇔ = π + π
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⇔ ( )
k
x k
4 2
π π
= + ∈ Z
so vôùi ñieàu kieän
Caùch 1 : Khi
k
x thì
4 2
π π
= + ( )
3 3k 2
cos 3x cos 0 nhaän
4 2 2
π π⎛ ⎞
= + = ± ≠⎜ ⎟
⎝ ⎠
Caùch 2 : Bieåu dieãn caùc ngoïn cung ñieàu kieän vaø ngoïn cung nghieäm ta thaáy
khoâng coù ngoïn cung naøo truøng nhau. Do ñoù :
(*) ⇔
k
x
4 2
π π
= +
Löu yù caùch 2 raát maát thôøi gian
Caùch 3 :
Neáu
π π π
= + = +
3 3k
πh
4 2 2
h 6k
3x
Thì + = +3 6k 2 4h
⇔1 4= −
⇔ = −
1
2h 3k
2
(voâ lyù vì ∈k, )h Z
( )2 2 2 11
tg x cot g x cot g 2x *
3
+ + =
Ñieàu kieän
cos x 0
sin x 0 sin 2x 0
sin 2x 0
≠⎧
⎪
≠ ⇔ ≠⎨
⎪ ≠⎩
Do ñoù :
(*)⇔ 2 2 2
1 1 1
1 1 1
cos x sin x sin 2x 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
− + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
11
⇔ 2 2 2 2
1 1 1
cos x sin x 4 sin x cos x 3
+ + =
20
⇔
2 2
2 2
4 sin x 4 cos x 1 20
4 sin x cos x 3
+ +
=
⇔ 2
5 2
sin 2x 3
=
0
⇔ 2 3
sin 2x
4
= (nhaän do sin2x 0≠ )
⇔ ( )
1 3
1 cos 4x
2 4
− =
⇔
1 2
cos 4x cos
2 3
π
= − =
⇔
2
4x k2
3
π
= ± + π
⇔ ( )
k
x k
6 2
π π
= ± + ∈ Z
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Chuù yù : Coù theå deã daøng chöùng minh :
2
tgx cot gx
sin 2x
+ =
Vaäy (*)⇔ ( )
2
2
1 1
tgx cot gx 2 1
sin x 3
⎛ ⎞
+ − + − =⎜ ⎟
⎝ ⎠
1
⇔ 2
5 2
sin 2x 3
=
0
Giaûi phöông trình
( )2 2 2x x
sin tg x cos 0 *
2 4 2
π⎛ ⎞
− − =⎜ ⎟
⎝ ⎠
Ñieàu kieän : cos x 0 sin x 1≠ ⇔ ≠ ±
luùc ñoù :
(*) ⇔ [ ]
2
2
1 sin x 1
1 cos x 1 cos x 0
2 2 cos x 2
⎡ π ⎤⎛ ⎞
− − − +⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
=
⇔
( )( )
( )
2
2
1 sin x 1 cos x
1 cos x 0
1 sin x
− −
− + =
−
⇔ ( )
2
1 cos x
1 cos x 0
1 sin x
−
− + =
+
⇔ ( )
1 cos x
1 cos x 1 0
1 sin x
−⎡ ⎤
+ −⎢ ⎥+⎣ ⎦
=
=⇔ ( )( )1 cos x cos x sin x 0+ − −
⇔
( )cos x 1 nhaändocos x 0
tgx 1
= − ≠⎡
⎢
= −⎣
⇔
= π + π⎡
⎢ π⎢ = − + π
⎣
x k2
x k
4
( ) ( )2
sin 2x cot gx tg2x 4 cos x *+ =
Ñieàu kieän : ⇔
sin x 0
cos2x 0
≠⎧
⎨
≠⎩
2
sin x 0
2cos x 1 0
≠⎧
⎨
− ≠⎩
⇔
cos x 1
2
cos x
2
≠ ±⎧
⎪
⎨
≠ ±⎪
⎩
Ta coù :
cos x sin 2x
cot gx tg2x
sin x cos2x
+ = +
cos2x cos x sin 2xsin x
sin x cos2x
+
=
cos x
sin x cos2x
=
Luùc ñoù : (*)
⎛ ⎞
⇔ =⎜ ⎟
⎝ ⎠
2cos x
2sin x cos x 4 cos x
sin x cos 2x
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⇔
2
22cos x
4cos x
cos2x
= ( )Dosin x 0≠
⇔
cos x 0
1
2
cos2x
=⎡
⎢
⎢ =
⎣
⇔
( )
⎡ ⎛ ⎞
= ≠ ≠ ±⎢ ⎜ ⎟⎜ ⎟
⎢ ⎝ ⎠
⎢ π
⎢ = = ≠
⎣
2
cos x 0 Nhaän do cos x vaø 1
2
1
cos 2x cos , nhaän do sin x 0
2 3
⇔
π⎡
= + π⎢
⎢
π⎢ = ± + π
⎢⎣
x k
2
x k
6
( )∈k Z
Baøi 47 : Giaûi phöông trình:
( )
2 2
cot g x tg x
16 1 cos4x
cos2x
−
= +
Ta coù :
2 2
2 2
2 2
cos x sin x
cot g x tg x
sin x cos x
− = −
4 4
2 2 2
cos x sin x 4cos2x
sin x cos x sin 2x
−
= =
Ñieàu kieän : ⇔ si
sin 2x 0
cos2x 0
≠⎧
⎨
≠⎩
n4x 0≠
Luùc ñoù (*) ( )2
4
16 1 cos4x
sin 2x
⇔ = +
( )
( )( )
( )
( )
( )
⇔ = +
⇔ = + −
⇔ = − =
⇔ = ≠
⇔ − =
π π
⇔ = ⇔ = + ∈
2
2 2
2
1 4 1 cos 4x sin 2x
1 2 1 cos 4x 1 cos 4x
1 2 1 cos 4x 2sin 4x
1
sin 4x nhaän do sin 4x 0
2
1 1
1 cos 8x
2 2
k
cos 8x 0 x , k
16 8
Baøi 48: Giaûi phöông trình: ( )4 4 7
sin x cos x cot g x cot g x *
8 3 6
π π⎛ ⎞ ⎛ ⎞
+ = + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Ñieàu kieän
sin x 0 sin x 0
3 3 2
sin 2x 0
3
sin x 0 cos x 0
6 3
⎧ ⎧π π⎛ ⎞ ⎛ ⎞
+ ≠ + ≠⎜ ⎟ ⎜ ⎟⎪ ⎪
π⎪ ⎝ ⎠ ⎪ ⎝ ⎠ ⎛ ⎞
⇔ ⇔ +⎨ ⎨ ⎜ ⎟
π π ⎝ ⎠⎛ ⎞ ⎛ ⎞⎪ ⎪− ≠ + ≠⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎩
≠
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1 3
sin 2x cos2x 0
2 2
tg2x 3
⇔ − + ≠
⇔ ≠
Ta coù: ( )
2
4 4 2 2 2 2 21
sin x cos x sin x cos x 2sin x.cos x 1 sin 2x
2
+ = + − = −
Vaø: cot g x .cot g x cot g x .tg x 1
3 6 3 3
π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ − = + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=
Luùc ñoù: (*) 21 7
1 sin 2x
2 8
⇔ − =
( )
1 1
1 cos4x
4 8
⇔ − − = −
⇔ =
π π
⇔ = ± + π ⇔ = ± +
1
cos 4x
2
k
4x k2 x
3 1
π
2 2
(nhaän do
3
tg2x 3
3
= ± ≠ )
Baøi 49: Giaûi phöông trình ( )
1
2tgx cot g2x 2sin 2x *
sin 2x
+ = +
Ñieàu kieän:
cos2x 0
sin 2x 0 cos2x 1
sin 2x 0
≠⎧
⇔ ≠ ⇔ ≠ ±⎨
≠⎩
Luùc ñoù: (*)
2sin x cos2x 1
2sin 2x
cos x sin 2x sin 2x
⇔ + = +
( )
( )
( )
( )
( )
( )
⇔ + = +
⇔ + − = +
⇔ − =
⎡ ⎤⇔ − + =⎣ ⎦
⎡ = ≠ ⇒
⎢⇔ π⎢
≠
= − = ≠ ±
⎢⎣
π
⇔ = ± + π ∈
π
⇔ = ± + π ∈
2 2
2 2 2 2
2 2
2
4 sin x cos 2x 2sin 2x 1
4 sin x 1 2sin x 8sin x cos x 1
2sin x 1 4 cos x 0
2sin x 1 2 1 cos 2x 0
sin x 0 loaïi do sin 2x 0 sin x 0
1 2
cos 2x cos nhaän do cos 2x 1
2 3
2
2x k2 k Z
3
x k , k
3
Baøi 51: Giaûi phöông trình:
( )
( )
3 sin x tgx
2 1 cos x 0 *
tgx sin x
+
− + =
−
( )
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Ñieàu kieän : ⇔tgx sin x 0− ≠
sin x
sin x 0
cos x
− ≠
⇔
( )sin x 1 cos x
0
cos x
−
≠ ⇔
sin x 0
cos x 0 sin 2x 0
cos x 1
≠⎧
⎪
≠ ⇔ ≠⎨
⎪ ≠⎩
Luùc ñoù (*)⇔
( )
( )
( )
3 sin x tgx .cot gx
2 1 cos x 0
tgx sin x .cot gx
+
− + =
−
⇔
( )
( )
( )
3 cos x 1
2 1 cos x 0
1 cos x
+
− + =
−
⇔ ( )− = ≠ + ≠
−
3
2 0 do sin x 0 neân cos x 1 0
1 cos x
⇔ 1 2cos x 0+ =
⇔
1
cos x
2
= − (nhaän so vôùi ñieàu kieän)
⇔
π
= ± + π ∈
2
x k2 , k
3
( ) ( )
( )
( ) ( )
2 2
2 21 cos x 1 cos x 1
tg xsin x 1 sin x tg x *
4 1 sin x 2
− + +
− = + +
−
Ñieàu kieän :
cos x 0
sin x 1
≠⎧
⎨
≠⎩
⇔ cos x 0≠
Luùc ñoù (*)⇔
( )
( )
( )
2 3 2
2 2
2 1 cos x sin x 1 sin x
1 sin x
4 1 sin x 1 sin x 2 1 sin x
+
− = + +
− − −
⇔ ( )( ) ( )( )2 3 2
1 cos x 1 sin x 2sin x 1 sin x 1 sin x 2sin x+ + − = + − + 2
⇔ ( )( ) ( ) ( )2 2 2
1 sinx 1 cos x 1 sin x cos x 2sin x 1 sin x+ + = + + +
⇔ 2 2
1 sin x 0
1 cos x cos x 2sin x
+ =⎡
⎢
+ = +⎣
2
⇔ ⇔ cos2x = 0
= − ≠⎡
⎢ = −⎣
sin x 1( loaïi do cos x 0)
1 1 cos 2x
⇔ 2x k
2
π
= + π
⇔ x k
4 2
π
= +
π
(nhaän do cosx ≠ 0)
( )cos3x.tg5x sin7x *=
Ñieàu kieän cos5x 0≠
Luùc ñoù : (*) ⇔
sin5x
cos3x. sin7x
cos5x
=
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⇔ sin5x.cos3x sin7x.cos5x=
⇔ [ ] [ ]
1 1
sin8x sin 2x sin12x sin2x
2 2
+ = +
⇔ sin8x sin12x=
⇔ 12x 8x k2 12x 8x k2= + π ∨ = π − + π
⇔
π π
= ∨ = +
k k
x x
2 20
π
10
So laïi vôùi ñieàu kieän
k 5k
x thì cos5x cos cos
2 2
π π
= = =
k
2
π
(loaïi neáu k leû)
π π π π⎛ ⎞
= + = + ≠⎜ ⎟
⎝ ⎠
k k
x thì cos5x cos 0
20 10 4 2
nhaän
Do ñoù : (*)⇔
π π
= π ∨ = +
k
x h x
20 10
, vôùi k, h ∈
( )
4 4
sin x cos x 1
tgx cot g2x *
sin2x 2
+
= + ( )
Ñieàu kieän : sin2x 0≠
Ta coù : ( )
24 4 2 2 2
sin x cos x sin x cos x 2sin x cos x+ = + − 2
21
1 sin 2
2
= − x
sin x cos2x
tgx cot g2x
cos x sin2x
+ = +
sin2xsin x cos x cos2x
cos xsin2x
+
=
( )cos 2x x 1
cos xsin 2x sin 2x
−
= =
( )
−
⇔ =
⇔ − =
⇔ = ≠
⇔ =
π
⇔ = + π ∈
π π
⇔ = + ∈
2
2
2
2
1
1 sin 2x
12Do ñoù : (*)
sin 2x 2sin 2x
1 1
1 sin 2x
2 2
sin 2x 1 nhaän do sin 2x 0
cos 2x 0
2x k , k
2
k
x , k
4 2
( )2 2 2 2
tg x.cot g 2x.cot g3x tg x cot g 2x cot g3x *= − +
Ñieàu kieän : cos x 0 sin2x 0 sin3x 0≠ ∧ ≠ ∧ ≠
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sin2x 0 sin3x 0⇔ ≠ ∧ ≠
( )⇔ − = −
⎡ − + ⎤ − +⎛ ⎞ ⎛ ⎞
⇔ − =⎜ ⎟ ⎜ ⎟⎢ ⎥+ − + −⎝ ⎠ ⎝ ⎠⎣ ⎦
2 2 2 2
Luùc ñoù (*) cotg3x tg x cot g 2x 1 tg x cot g 2x
1 cos 2x 1 cos 4x 1 cos 2x 1 cos 4x
cot g3x 1
1 cos 2x 1 cos 4x 1 cos 2x 1 cos 4x
−
( )( ) ( )( )
( )( ) ( )( )
[ ] ( )
[ ]
( )
⎡ ⎤⇔ − + − + −⎣ ⎦
= − − − + +
⇔ − = − +
⇔ − = −
⇔ = ≠
⇔ = ∨ =
π
⇔ = + π ∨ =
cot g3x 1 cos2x 1 cos4x 1 cos2x 1 cos4x
cot g3x 2cos4x 2cos2x 2 cos4x cos2x
cos3x
4sin 3xsin x 4cos3x cos x
sin 3x
cos3xsin x cos3x cos x do sin 3x 0
cos3x 0 sin x cos x
3x k tgx 1
2
( )
π π π
⇔ = + ∨ = + π ∈
k
x x l k,l Z
6 3 4
So vôùi ñieàu kieän: ≠sin2x.sin3x 0
* Khi
π π
= +
k
x
6 3
thì
π π π⎛ ⎞ ⎛ ⎞
+ + π ≠⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2k
sin .sin k 0
3 3 2
+⎛ ⎞
⇔ π⎜ ⎟
⎝ ⎠
1 2k
sin 0
3
≠
Luoân ñuùng ( )∀ + ≠k thoûa 2k 1 3m m Z∈
* Khi
π
= +x l
4
π thì
π π⎛ ⎞ ⎛ ⎞
+ π + π = ±⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3 2
sin 2l sin 3l 0
2 4 2
≠
luoân ñuùng
Do ñoù: (*)
π π⎡
= + ∈ ∧ ≠ − ∈⎢
⇔ ⎢
π⎢ = + π ∈
⎢⎣
k
x , k Z 2k 3m 1( m
6 3
x l , l
4
)
Caùch khaùc:
( )⇔ − = −
−−
⇔ = =
− −
+ −
⇔ =
− +
⇔ = ⇔ = ∨ =
2 2 2 2
2 22 2
2 2 2 2
(*) cotg3x tg x cot g 2x 1 tg x cot g 2x
tg 2x.tg x 1tg x cot g 2x
cot g3x
tg x cot g 2x 1 tg x tg 2x
(1 tg2x.tgx ) (1 tg2x.tgx )
cot g3x
(tg2x tgx) ( tg2x tgx)
cot g3x cot gx.cotg3x cos 3x 0 sin x cos x
BAØI TAÄP
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1. Tìm caùc nghieäm treân
π⎛ ⎞
π⎜
⎝ ⎠
,3
3
⎟ cuûa phöông trình:
π π⎛ ⎞ ⎛ ⎞
+ − − = +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
5 7
sin 2x 3cos x 1 2sin x
2 2
2. Tìm caùc nghieäm x treân
π⎛
⎜
⎝ ⎠
0,
2
⎞
⎟ cuûa phöông trình
( )− = π +2 2
sin 4x cos 6x sin 10,5 10x
3. Giaûi caùc phöông trình sau:
a/ ( )3 3 5 5
sin x cos x 2 sin x cos x+ = +
b/
sin x sin 2x sin 3x
3
cos x cos2x cos3x
+ +
=
+ +
c/ 2 1 cos x
tg x
1 sin x
+
=
−
d/ tg2x tg3x tg5x tg2x.tg3x.tg5x− − =
e/ 24
cos x cos x
3
=
f/
1 1
2 2 sin x
4 sin x cos
π⎛ ⎞
+ = +⎜ ⎟
⎝ ⎠ x
i/
2
2tgx cot g2x 3
sin2x
+ = +
h/
2
3tg3x cot g2x 2tgx
sin 4x
+ = +
k/ 2 2 2
sin x sin 2x sin 3x 2+ + =
l/
sin2x
2cos x 0
1 sin x
+ =
+
m/ ( )2
25 4x 3sin 2 x 8sin x 0− π + π =
n/
sin x.cot g5x
1
cos9x
=
o/
2
3tg6x 2tg2x cot g4x
sin8x
− = −
p/ ( )2
2sin 3x 1 4sin x 1− =
q/ 2 1 cos x
tg x
1 sin x
+
=
−
r/ 3 3 2
cos x cos3x sin xsin 3x
4
+ =
s/ 4 4x x
sin cos
3 3
⎛ ⎞ ⎛ ⎞ 5
8
+ =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
t/ 3 3 2
cos x 4sin x 3cos xsin x sin x 0− − + =
u/ 4 4x x
sin cos 1 2sin x
2 2
+ = −
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v/ sin 3x sin 2x.sin x
4 4
π π⎛ ⎞ ⎛
− = +⎜ ⎟ ⎜
⎝ ⎠ ⎝
⎞
⎟
⎠
w/
( )2
4
4
2 sin x sin 3x
tg x 1
cos x
−
+ =
y/ 2 x
tgx cos x cos x sin x 1 tg tgx
2
⎛ ⎞
+ − = +⎜ ⎟
⎝ ⎠
4. Cho phöông trình: ( )( ) (2
2sin x 1 2cos2x 2sin x m 3 4 cos x 1− + + = − )
a/ Giaûi phöông trình khi m = 1
b/ Tìm m ñeå (1) coù ñuùng 2 nghieäm treân [ ]0,π
( ÑS: m 0 m 1 m 3= ∨ < − ∨ > )
5. Cho phöông trình:
( )5 5 2
4cos xsin x 4sin x.cos x sin 4x m 1− = +
Bieát raèng x = π laø moät nghieäm cuûa (1). Haõy giaûi phöông trình trong tröôøng
hôïp ñoù.
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CHÖÔNG III.
PHÖÔNG TRÌNH BAÄC HAI VÔÙI CAÙC HAØM SOÁ LÖÔÏNG GIAÙC
( )
( )
( )
( )
+ + = ≠
+ + = ≠
+ = = ≠
+ + =
2
2
2
2
a sin u b sin u c 0 a 0
a cos u b cos u c 0 a 0
atg u btgu c 0 a 0
a cot g u b cot gu c 0 a 0≠
Caùch giaûi:
Ñaët : hay vôùit sinu= t cosu= t 1≤
(ñieàu kieänt tgu= u k
2
π
≠ + π )
(ñieàu kieänt cot gu= u k≠ π )
Caùc phöông trình treân thaønh: 2
at bt c 0+ + =
Giaûi phöông trình tìm ñöôïc t, so vôùi ñieàu kieän ñeå nhaän nghieäm t.
Töø ñoù giaûi phöông trình löôïng giaùc cô baûn tìm ñöôïc u.
Baøi 56: (Ñeà thi tuyeån sinh Ñaïi hoïc khoái A, naêm 2002)
Tìm caùc nghieäm treân cuûa phöông trình(0,2π)
( )
cos3x sin 3x
5 sin x 3 cos2x *
1 2sin 2x
+⎛ ⎞
+ = +⎜ ⎟
+⎝ ⎠
Ñieàu kieän:
1
sin 2x
2
≠ −
Ta coù: ( ) ( )3 3
sin 3x cos3x 3sin x 4sin x 4 cos x 3cos x+ = − + −
( ) ( )
( ) ( )
( )( )
3 3
2 2
3 cos x sin x 4 cos x sin x
cos x sin x 3 4 cos x cos xsin x sin x
cos x sin x 1 2sin 2x
= − − + −
⎡ ⎤= − − + + +
⎣ ⎦
= − +
Luùc ñoù: (*) ( ) ( )2
5 sin x cos x sin x 3 2cos x 1⎡ ⎤⇔ + − = +⎣ ⎦ −
1
do sin 2x
2
⎛ ⎞
≠ −⎜ ⎟
⎝ ⎠
2
2cos x 5cos x 2 0⇔ − + =
( )
1
cos x
2
cos x 2 loaïi
⎡
=⎢⇔
⎢
=⎢⎣
x
3
π
⇔ = ± + πk2 (nhaän do
3 1
sin 2x
2 2
= ± ≠ − )
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Do ( )x 0,2∈ π neân
5
x x
3 3
π π
= ∨ =
Baøi 57: (Ñeà thi tuyeån sinh Ñaïi hoïc khoái A, naêm 2005)
Giaûi phöông trình: ( )2 2
cos 3x.cos2x cos x 0 *− =
Ta coù: (*)
1 cos6x 1 cos2x
.cos2x 0
2 2
+ +
⇔ − =
cos6x.cos2x 1 0⇔ − = (**)
Caùch 1: (**) ( )3
4 cos 2x 3cos2x cos2x 1 0⇔ − − =
=4 2
4 cos 2x 3cos 2x 1 0⇔ − −
( )
2
2
cos 2x 1
1
cos 2x voâ nghieäm
4
⎡ =
⎢⇔
⎢ = −
⎢⎣
( )
sin 2x 0
k
2x k x k Z
2
⇔ =
π
⇔ = π ⇔ = ∈
Caùch 2: (**) ( )
1
cos8x cos4x 1 0
2
⇔ + − =
( )
2
cos8x cos4x 2 0
2cos 4x cos4x 3 0
cos4x 1
3
cos4x loaïi
2
⇔ + − =
⇔ + −
=⎡
⎢⇔
⎢ = −
⎣
=
( )
k
4x k2 x k Z
2
π
⇔ = π ⇔ = ∈
Caùch 3: phöông trình löôïng giaùc khoâng maãu möïc:
(**)
cos6x cos2x 1
cos6x cos2x 1
= =⎡
⇔ ⎢ = = −⎣
Caùch 4: + − = ⇔ +cos 8x cos 4x 2 0 cos 8x cos 4x 2=
⇔ = =cos 8x cos 4x 1 ⇔ =cos 4x 1
Baøi 58: (Ñeà thi tuyeån sinh Ñaïi hoïc khoái D, naêm 2005)
Giaûi phöông trình: 4 4 3
cos x sin x cos x sin 3x 0
4 4
π π⎛ ⎞ ⎛ ⎞
+ + − − −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ 2
=
Ta coù:
(*)
( )
22 2 2 2 1 3
sin x cos x 2sin x cos x sin 4x sin 2x 0
2 2
⎡ ⎤π⎛ ⎞
⇔ + − + − + −⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦ 2
=
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[ ]21 1 3
1 sin 2x cos4x sin 2x 0
2 2 2
⇔ − + − + − =
( )2 21 1 1 1
sin 2x 1 2sin 2x sin2x 0
2 2 2 2
⇔ − − − + − =
2
sin 2x sin 2x 2 0⇔ + − =
( )
sin 2x 1
sin 2x 2 loaïi
=⎡
⇔ ⎢
= −⎣
π
⇔ = + π ∈
π
⇔ = + π ∈
2x k2 , k
2
x k , k
4
Baøi 59: (Ñeà thi tuyeån sinh Ñaïi hoïc khoái B, naêm 2004)
Giaûi phöông trình: ( ) (− = − 2
5sin x 2 3 1 sinx tg x *)
Ñieàu kieän: cos x 0 sin x 1≠ ⇔ ≠ ±
Khi ñoù: (*) ( )
2
2
sin x
5sin x 2 3 1 sin x
cos x
⇔ − = −
( )
2
2
sin x
5sin x 2 3 1 sin x
1 sin x
⇔ − = −
−
2
3sin x
5sin x 2
1 sin x
⇔ − =
+
2
2sin x 3sin x 2 0⇔ + − =
( )
( )
1
sin x nhaändosin x 1
2
sin x 2 voâ nghieäm
⎡
= ≠⎢⇔
⎢
= −⎢⎣
±
( )
5
x k2 x k2 k
6 6
π π
⇔ = + π ∨ = + π ∈ Z
Baøi 60: Giaûi phöông trình: ( )
1 1
2sin 3x 2cos3x *
sin x cos x
− = +
Ñieàu kieän: sin2x 0≠
Luùc ñoù: (*) ( )
1 1
2 sin 3x cos3x
sin x cos x
⇔ − = +
( ) ( )3 3 1 1
2 3 sin x cos x 4 sin x cos x
sin x cos x
⎡ ⎤⇔ + − + = +
⎣ ⎦
( ) ( )2 2 sin x cos x
2 sin x cos x 3 4 sin x sin x cos x cos x
sin x cos x
+⎡ ⎤⇔ + − − + =
⎣ ⎦
( )
1
sin x cos x 2 8sin x cos x 0
sin x cos x
⎡ ⎤
⇔ + − + −⎢ ⎥⎣ ⎦
=
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( )
2
sin x cos x 4sin 2x 2 0
sin 2x
⎡ ⎤
⇔ + − −⎢ ⎥⎣ ⎦
=
( )2
tgx 1sin x cos x 0
nhaän so vôùiñieàu kieän1
sin 2x 1 sin 2x4sin 2x 2sin 2x 2 0
2
= −⎡+ =⎡ ⎢⇔ ⇔ −⎢ ⎢ = ∨ =− − =⎣ ⎣
π π π π
⇔ = − + π ∨ = + π ∨ = − + π ∨ = + π ∈
7
x k 2x k2 2x k2 2x k2 , k
4 2 6 6
π π π
⇔ = ± + π ∨ = − + π ∨ = + π ∈
7
x k x k x k , k
4 12 12
( ) ( )
+ − −
=
+
2
cos x 2sin x 3 2 2 cos x 1
1 *
1 sin 2x
Ñieàu kieän: sin 2x 1 x m
4
π
≠ − ⇔ ≠ − + π
Luùc ñoù:
(*) 2
2sin x cos x 3 2 cos x 2cos x 1 1 sin 2x⇔ + − − = +
2
2cos x 3 2 cos x 2 0⇔ − + =
( )⇔ = =
2
cos x hay cos x 2 voâ nghieäm
2
( )
x k2
4
x k '2 loaïi do ñieàu ki
4
π⎡
= + π⎢
⇔ ⎢
π⎢ = − + π
⎢⎣
eän
x k
4
π
⇔ = + π2
( )
x 3x x 3x 1
cos x.cos .cos sin xsin sin *
2 2 2 2 2
− =
Ta coù: (*) ( ) ( )
1 1
cos x cos2x cos x sin x cos2x cos x
2 2
1
2
⇔ + + − =
2
cos x.cos2x cos x sin x cos2x sin x cos x 1⇔ + + − =
( ) 2
cos2x cos x sin x 1 cos x sin x cos x⇔ + = − +
( ) ( )cos2x cos x sin x sin x sin x cos x⇔ + = +
( )( ) ( )cos x sin x cos2x sin x 0 * *⇔ + − =
( )( )2
cos x sin x 1 2sin x sin x 0⇔ + − − =
2
cos x sin x
2sin x sin x 1 0
= −⎡
⇔ ⎢
+ − =⎣
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tgx 1
sin x 1
1
sin x
2
⎡
⎢ = −
⎢
⇔ =⎢
⎢
=⎢
⎣
− ( )
x k
4
x k2 k
2
5
x k2 x k2
6 6
π⎡
= − + π⎢
⎢
π⎢⇔ = − + π ∈
⎢
⎢ π π⎢ = + π ∨ = + π
⎢⎣
Z
Caùch khaùc: (**) tgx 1 cos2x sin x cos x
2
π⎛ ⎞
⇔ = − ∨ = = −⎜ ⎟
⎝ ⎠
Baøi 63: Giaûi phöông trình: ( )3
4 cos x 3 2 sin 2x 8cos x *+ =
Ta coù: (*) 3
4 cos x 6 2 sin x cos x 8cos x 0⇔ + − =
( )2
cos x 2cos x 3 2 sin x 4 0⇔ + − =
( )2
cos x 2 1 sin x 3 2 sin x 4 0⎡ ⎤⇔ − + −
⎣ ⎦
=
2
cos x 0 2sin x 3 2 sin x 2 0⇔ = ∨ − + =
( )
cos x 0
2
sin x
2
sin x 2 voâ nghieäm
=⎡
⎢
⎢⇔ =
⎢
⎢
=⎢⎣
2
x k sin x sin
2 2
π π
⇔ = + π ∨ = =
4
( )
3
x k x k2 x k2 k
2 4 4
π π π
⇔ = + π ∨ = + π ∨ = + π ∈ Z
( )cos 2x cos 2x 4sin x 2 2 1 sin x *
4 4
π π⎛ ⎞ ⎛ ⎞
+ + − + = + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( )
(*) ( )2cos2x.cos 4sin x 2 2 1 sin x
4
π
⇔ + = + −
( ) ( )
( )
2
2
2 1 2sin x 4 2 sin x 2 2 0
2 2 sin x 4 2 sin x 2 0
⇔ − + + − − =
⇔ − + + =
( )⇔ − + + =2
2sin x 2 2 1 sin x 2 0
( )⎡ =
⎢⇔ ⎢ =⎢⎣
sin x 2 loaïi
1
sin x
2
π π
⇔ = + π = + π ∈
5
x k2 hay x k2 , k
6 6
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Baøi 65: Giaûi phöông trình : ( ) ( )+ = +2 2
3cot g x 2 2 sin x 2 3 2 cos x *
Ñieàu kieän: sin x 0 cos x 1≠ ⇔ ≠ ±
Chia hai veá (*) cho ta ñöôïc:2
sin x
(*) ( )
2
4 2
cos x cos x
3 2 2 2 3 2
sin x sin x
⇔ + = + vaø sin x 0≠
Ñaët 2
cos x
t
sin x
= ta ñöôïc phöông trình:
( )2
3t 2 3 2 t 2 2 0
2
t 2 t
3
− + + =
⇔ = ∨ =
* Vôùi
2
t
3
= ta coù: 2
cos x 2
3sin x
=
( )
( )
( )
2
2
3cos x 2 1 cos x
2cos x 3cos x 2 0
cos x 2 loaïi
1
cos x nhaän do cos x 1
2
⇔ = −
⇔ + − =
⎡ = −
⎢⇔
⎢ = ≠
⎢⎣
±
( )x k2 k
3
π
⇔ = ± + π ∈ Z
* Vôùi t = 2 ta coù: =2
cos x
2
sin x
( )
( )
( )
⇔ = −
⇔ + − =
⎡ = −
⎢
⇔ ⎢
= ≠ ±⎢
⎣
π
⇔ = ± + π ∈
2
2
cos x 2 1 cos x
2 cos x cos x 2 0
cos x 2 loaïi
2
2
x k2 , k
4
Baøi 66: Giaûi phöông trình: ( )
+ − −
=
2 2
4 sin 2x 6sin x 9 3cos 2x
0 *
cos x
Ñieàu kieän: ≠cos x 0
Luùc ñoù:
(*) 2 2
4sin 2x 6sin x 9 3cos2x 0⇔ + − − =
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( ) ( )2
2
4 1 cos 2x 3 1 cos2x 9 3cos2x 0
4 cos 2x 6cos2x 2 0
1
cos2x 1 cos2x
2
⇔ − + − − − =
⇔ + + =
⇔ = − ∨ = −
2 2 1
2cos x 1 1 2cos x 1
2
⇔ − = − ∨ − = −
( )
( )
( )
cos x 0 loaïi doñieàu kieän
1
2
2
x k2 x k2 k
3 3
⎡ =
⎢⇔ ⎢ = ± ≠⎢⎣
π π
⇔ = ± + π ∨ = ± + π ∈ Z
Baøi 67: Cho ( )
1 2
f x sin x sin 3x sin5x
3 5
= + +
Giaûi phöông trình: ( )f ' x 0=
Ta coù: ( )f ' x 0=
( ) ( )
( ) ( )3 2
cos x cos3x 2cos5x 0
cos x cos5x cos3x cos5x 0
2cos3x cos2x 2cos4x cos x 0
4 cos x 3cos x cos2x 2cos 2x 1 cos x 0
⇔ + + =
⇔ + + + =
⇔ + =
⇔ − + − =
( )
( )
⎡ ⎤⇔ − + −
⎣ ⎦
⎡⎡ ⎤+ − + − =⎣ ⎦⇔ ⎢
=⎢⎣
⎡ − − =
⇔ ⎢
=⎣
±
⇔ = ∨ =
2 2
2
2
4 cos x 3 cos 2x 2 cos 2x 1 cos x 0
2 1 cos 2x 3 cos 2x 2cos 2x 1 0
cos x 0
4 cos 2x cos 2x 1 0
cos x 0
1 17
cos 2x cos x 0
8
=
( )
1 17 1 17
cos2x cos cos2x cos cos x 0
8 8
x k x k x k k Z
2 2 2
+ −
⇔ = = α ∨ = = β ∨
α β π
⇔ = ± + π ∨ = ± + π ∨ = + π ∈
=
Baøi 68: Giaûi phöông trình: ( )8 8 217
sin x cos x cos 2x *
16
+ =
Ta coù:
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( )
( )
2
8 8 4 4 4 4
22
2 2 2 2 4
2
2 4
2 4
sin x cos x sin x cos x 2sin x cos x
1
sin x cos x 2sin x cos x sin 2x
8
1 1
1 sin 2x sin 2x
2 8
1
1 sin 2x sin 2x
8
+ = + −
⎡ ⎤= + − −⎢ ⎥⎣ ⎦
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
= − +
Do ñoù:
( ) ( )
( )
( )
( ) ( )
⎛ ⎞
⇔ − + = −⎜ ⎟
⎝ ⎠
⇔ + − =
⎡ = −
⎢⇔ ⇔ −⎢ =⎢⎣
π
⇔ = ⇔ = + ∈
2 4
4 2
2
2
1
* 16 1 sin 2x sin 2x 17 1 sin 2x
8
2sin 2x sin 2x 1 0
sin 2x 1 loaïi
1 1
1 cos 4x1 2 2sin 2x
2
cos 4x 0 x 2k 1 , k Z
8
=
2
Baøi 69: Giaûi phöông trình: ( )35x x
sin 5cos x.sin *
2 2
=
Nhaän xeùt thaáy:
x
cos 0 x k2 cos x 1
2
= ⇔ = π + π ⇔ = −
Thay vaøo (*) ta ñöôïc:
π⎛ ⎞ ⎛
+ π = − + π⎜ ⎟ ⎜
⎝ ⎠ ⎝
5
sin 5k 5.sin k
2 2
π ⎞
⎟
⎠
, khoâng thoûa k∀
Do
x
cos
2
khoâng laø nghieäm cuûa (*) neân:
( ) ⇔ = 25x x x x
* sin .cos 5cos x.sin cos
2 2 2 2
vaø
x
cos 0
2
≠
( ) 31 5
sin 3x sin 2x cos x.sin x
2 2
⇔ + = vaø ≠
x
cos 0
2
3 3
3sin x 4sin x 2sin x cos x 5cos x.sin x⇔ − + = vaø ≠
x
cos 0
2
2 3
x
cos 0
2
3 4sin x 2cos x 5cos x sin x 0
⎧
≠⎪
⇔ ⎨
⎪ − + = ∨⎩ =
3 2
x
cos 0
2
x
5cos x 4 cos x 2cos x 1 0 sin 0
2
⎧
≠⎪⎪
⇔ ⎨
⎪ − − + = ∨
⎪⎩
=
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( )( )2
cos x 1
x
cos x 1 5cos x cos x 1 0 sin 0
2
≠ −⎧
⎪
⇔ ⎨
− + − = ∨ =⎪⎩
≠ −⎧
⎪
⎡⎪
⎢ =⎪
⎢⎪
⇔ − +⎨⎢ = = α⎪⎢
⎪⎢
− −⎪⎢
= = β⎪⎢⎣⎩
cos x 1
cos x 1
1 21
cos x cos
10
1 21
cos x cos
10
( )⇔ = π = ±α + π = ±β + π ∈x k2 hay x k2 hay x k2 , k Z
Baøi 70: Giaûi phöông trình: ( ) ( )2
sin 2x cot gx tg2x 4 cos x *+ =
Ñieàu kieän: cos2x 0≠ vaø sin x 0 cos2x 0 cos2x 1≠ ⇔ ≠ ∧ ≠
Ta coù:
cos x sin 2x
cot gx tg2x
sin x cos2x
+ = +
cos2x cos x sin 2xsin x
sin x cos2x
cos x
sin x cos2x
+
=
=
Luùc ñoù: (*) 2cos x
2sin x.cos x 4 cos x
sin x cos2x
⎛ ⎞
⇔ =⎜ ⎟
⎝ ⎠
( ) ( )
( )
( )
⇔ =
⇔ + = +
⇔ + = =
⇔ = − ∨ = ≠ ≠
2
2cos x
2cos x
cos 2x
cos 2x 1 2cos 2x cos 2x 1
cos 2x 1 0 hay 1 2cos 2x
1
cos 2x 1 cos 2x nhaän do cos 2x 0 vaø cos 2x 1
2
π
⇔ = π + π ∨ = ± + π ∈
π π
⇔ = + π ∨ = ± + π ∈
2x k2 2x k2 , k
3
x k x k , k
2 6
Baøi 71: Giaûi phöông trình: ( )2 6x 8x
2cos 1 3cos *
5 5
+ =
Ta coù : (*)
⎛ ⎞ ⎛ ⎞
⇔ + + =⎜ ⎟ ⎜
⎝ ⎠ ⎝
212x 4x
1 cos 1 3 2cos 1
5 5
− ⎟
⎠
⎛ ⎞
⇔ + − = ⎜ ⎟
⎝ ⎠
3 24x 4x 4x
2 4cos 3cos 3 2cos 1
5 5 5
−
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Ñaët ( )4
t cos x ñieàu kieän t 1
5
= ≤
Ta coù phöông trình :
( )( )
( )
3 2
3 2
2
4t 3t 2 6t 3
4t 6t 3t 5 0
t 1 4t 2t 5 0
1 21 1 21
t 1 t t loïai
4 4
− + = −
⇔ − − + =
⇔ − − − =
− +
⇔ = ∨ = ∨ =
Vaäy
( )
• = ⇔ = π
π
⇔ = ∈
4x 4x
cos 1 2k
5 5
5k
x k
2
Z
( )
( )
4x 1 21
cos cos vôùi 0 2
5 4
4x
2
5
5 5
x , Z
4 2
−
• = = α < α <
⇔ = ±α + π
α π
⇔ = ± + ∈
π
tg x tgx 1 *
4
π⎛ ⎞
− = −⎜ ⎟
⎝ ⎠
Ñaët t x x t
4 4
π π
= − ⇔ = +
(*) thaønh : 3 1 tgt
tg t tg t 1 1 vôùi cost 0 tgt 1
4 1 tgt
π +⎛ ⎞
= + − = − ≠ ∧ ≠⎜ ⎟ −⎝ ⎠
⇔ =
−
3 2tgt
tg t
1 tgt
( )
( )( )
( )
3 4
3 2
2
tg t tg t 2tgt
tgt tg t tg t 2 0
tgt tgt 1 tg t 2tgt 2 0
tgt 0 tgt 1 nhaän so ñieàu kieän
t k t k ,k
4
⇔ − =
⇔ − + =
⇔ + − + =
⇔ = ∨ = −
π
⇔ = π ∨ = − + π ∈
Vaäy (*)
x k hay x k ,k
4
π
⇔ = + π = π ∈
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4 4
4sin 2x cos 2x
cos 4x (*)
tg x tg x
4 4
+
=
π π⎛ ⎞ ⎛ ⎞
− +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Ñieàu kieän
sin x cos x 0 sin 2x 0
4 4 2
sin x cos x 0 sin 2x 0
4 4 2
⎧ ⎧π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞
− − ≠ −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪
⎪ ⎝ ⎠ ⎝ ⎠ ⎪ ⎝ ⎠
⇔⎨ ⎨
π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪+ + ≠ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎩
≠
≠
±cos2x 0 sin2x 1⇔ ≠ ⇔ ≠
Do :
1 tgx 1 tgx
tg x tg x . 1
4 4 1 tgx 1 tgx
π π − +⎛ ⎞ ⎛ ⎞
− + =⎜ ⎟ ⎜ ⎟
+ −⎝ ⎠ ⎝ ⎠
=
Khi cos2x 0 thì :≠
( )
( )
( )
( )
4 4 4
2 2 4
2 4
2 4
4 2
2
2
2
* sin 2x cos 2x cos 4x
1 2sin 2xcos 2x cos 4x
1
1 sin 4x cos 4x
2
1
1 1 cos 4x cos 4x
2
2cos 4x cos 4x 1 0
cos 4x 1
1 sin 4x 11
cos 4x voâ nghieäm
2
sin4x 0
2sin2xcos2x 0
sin2x 0 do cos2x 0
2x k ,k x k
⇔ + =
⇔ − =
⇔ − =
⇔ − − =
⇔ − − =
⎡ =
⎢⇔ ⇔
⎢ = −
⎢⎣
⇔ =
⇔ =
⇔ = ≠
⇔ = π ∈ ⇔ =
− =
, k
2
π
∈
Baøi 74 :Giaûi phöông trình: ( )4 2
1 2
48 1 cot g2xcotgx 0 *
cos x sin x
− − + = ( )
Ta coù :
( )2 2
cos2x cosx
1 cot g2xcot gx 1 .
sin2x sinx
sin2xsinx cos2x cosx
sin xsin2x
cosx 1
do cosx 0
2sin xcosx 2sin x
+ = +
+
=
= = ≠
Luùc ñoù (*) 4 4
1 1
48 0
cos x sin x
⇔ − − =
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4 4
4 4 4 4
1 1 sin x cos
48
cos x sin x sin xcos x
+
⇔ = + =
x
4 4 4 4
4 2 2
4 2
48sin xcos x sin x cos x
3sin 2x 1 2sin xcos x
1
3sin 2x sin 2x 1 0
2
⇔ = +
⇔ = −
⇔ + − =
( )
( )
2
2
2
sin x loïai
3
1
sin x nhaän do 0
2
⎡
= −⎢
⇔ ⎢
⎢ = ≠
⎢⎣
( )
( )
1 1
1 cos4x
2 2
cos4x 0
4x k
2
k
x k
8 4
⇔ − =
⇔ =
π
⇔ = + π
π π
⇔ = + ∈ Z
( ) ( )8 8 10 10 5
sin x cos x 2 sin x cos x cos2x *
4
+ = + +
Ta coù : (*)
( ) ( )8 10 8 10 5
sin x 2sin x cos x 2cos x cos2x
4
⇔ − + − =
( ) ( )
( )
8 2 8 2
8 8
8 8
5
sin x 1 2sin x cos x 1 2cos x cos2x
4
5
sin x.cos2x cos xcos2x cos2x
4
4cos2x sin x cos x 5cos2x
⇔ − − − + =
⇔ − =
⇔ − =
( )
( )( )
8 8
4 4 4 4
2
2
cos2x 0 hay 4 sin x cos x 5
cos2x 0 hay 4 sin x cos x sin x cos x 5
1
cos2x 0 hay 4 1 sin 2x 5
2
cos2x 0 hay 2sin 2x 1(Voâ nghieäm)
⇔ = − =
⇔ = − +
⎛ ⎞
⇔ = − =⎜ ⎟
⎝ ⎠
⇔ = − =
=
2x k ,k
2
π
⇔ = + π ∈
k
x ,k
4 2
π π
⇔ = + ∈
Caùch khaùc: Ta coù ( )8 8
4 sin x cos x 5− = voâ nghieäm
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Vì ( )8 8
sin x cos x 1, x− ≤ ∀ neân ( )8 8
4 sin x cos x 4 5, x− ≤ < ∀
Ghi chuù : Khi gaëp phöông trình löôïng giaùc daïng R(tgx, cotgx, sin2x, cos2x, tg2x)
vôùi R haøm höõu tyû thì ñaët t = tgx
Luùc ñoù
2
2 2
2t 2t 1 t
tg2x ,sin2x ,cos2x
1 t 1 t 1 t2
−
= = =
− + +
Baøi 76 : (Ñeå thi tuyeån sinh Ñaïi hoïc khoái A, naêm 2003)
Giaûi phöông trình
( )− = + −
+
2cos 2x 1
cot gx 1 sin x sin 2x *
1 tgx 2
Ñieàu kieän : sin2x 0vaø tgx 1≠ ≠ −
Ñaët t = tgx thì (*) thaønh :
2
22
2 2
1 t
1 1 1 t 11 t1 1 .
t 1 t 2 1 t 2 1
−
⎡ ⎤−+− = + − −⎢ ⎥+ +⎣ ⎦
2t
t+
( )
( )
( )( ) ( )
( )
( )
( )
2
2 2 2
22
2 2
22
2 2
1 t 1 t 1 2t t
. do t
t 1 t 2 1 t 1 t
1 t1 t t 2t 1
t 1 t 1 t
1 t 1 t 1 t t
t 1 nhaän do t 11 t 0
1 t 1 t t 2t t 1 0 voâ nghieäm
− −
⇔ = + − ≠ −
+ + +
−− − +
⇔ = =
+ +
⇔ − + = −
= ≠ −⎡− =⎡
⇔ ⇔ ⎢⎢
+ = − − + =⎢⎣ ⎣
1
Vaäy (*) ⇔ ( )tgx 1 x k nhaän do sin2x 1 0
4
π
= ⇔ = + π = ≠
Baøi 77 : Giaûi phöông trình: ( )+ =sin 2x 2tgx 3 *
Ñieàu kieän : cos x 0≠
Ñaët t = tgx thì (*) thaønh :
2
2t
2t 3
1 t
+ =
+
( )( )
( )( )
( )
( )
⇔ + − + =
⇔ − + − =
⇔ − − + =
=⎡
⇔ ⎢
− + =⎣
π
⇔ = ⇔ = + π ∈
2
3 2
2
2
2t 2t 3 1 t 0
2t 3t 4t 3 0
t 1 2t t 3 0
t 1
2t t 3 0 voâ nghieäm
Vaäy (*) tgx 1 x k k Z
4
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( )
2
cot gx tgx 4sin2x *
sin 2x
− + =
Ñaët 2
2t
t tgx thì : sin 2x do sin 2x 0 neân t 0
1 t
= = ≠
+
≠
(*) thaønh :
2
2
1 8t 1 t 1
t t
t 1 t t t
+
− + = = +
+
( )
( )
⇔ =
+
⇔ = ≠
+
⇔ = ⇔ = ± ≠
π⎛ ⎞
⇔ = ±⎜ ⎟
⎝ ⎠
π
⇔ = ± + π ∈
2
2
2
8t
2t
1 t
4
1 do t 0
1 t
t 3 t 3 nhaän do t 0
Vaäy (*) tgx tg
3
x k , k
3
( )( ) ( )1 tgx 1 sin 2x 1 tgx *− + = +
Ñaët = tgx thì (*) thaønh :
( ) 2
2t
1 t 1 1 t
1 t
⎛ ⎞
− + = +⎜ ⎟
+⎝ ⎠
( )
( )
( )( )
2
2
2 2
2
t 1
1 t 1 t
1 t
t 1
t 1
1 t 1 t
1 t 1 t1
1 t
t 1 t 0
+
⇔ − = +
+
= −⎡
= −⎡⎢⇔ ⇔− + ⎢⎢ − = += ⎣⎢ +⎣
⇔ = − ∨ =
Do ñoù (*)
= −⎡ π
⇔ ⇔ = − + π = π ∈⎢ =⎣
tgx 1
x k hay x k , k
tgx 0 4
Baøi 80 : Cho phöông trình ( ) ( )cos2x 2m 1 cos x m 1 0 *− + + + =
a/ Giaûi phöông trình khi
3
m
2
=
b/ Tìm m ñeå (*) coù nghieäm treân
3
,
2 2
π π⎛ ⎞
⎜ ⎟
⎝ ⎠
Ta coù (*) ( )2
2cos x 2m 1 cos x m 0− + + =
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[ ]( )
( )
⎧ = ≤⎪
⇔ ⎨
− + + =⎪⎩
2
t cos x t 1
2t 2m 1 t m 0
[ ]( )⎧ = ≤
⎪
⇔ ⎨
= ∨ =⎪
⎩
t cos x t 1
1
t t m
2
a/ =
3
Khi m , phöông trình thaønh
2
b/
( )
( )
)
[ ]
( ) )
= ∨ =
π
⇔ = ± + π ∈
π π⎛ ⎞
∈ = ∈ −⎜ ⎟
⎝ ⎠
= ∉ −
π π⎛ ⎞
⇔ ∈ −⎡⎜ ⎟ ⎣
⎝ ⎠
1 3
cos x cos x loaïi
2 2
x k2 k Z
3
3
Khi x , thì cos x t [ 1, 0
2 2
1
Do t 1, 0 neân
2
3
* coù nghieäm treân , m 1, 0
2 2
( )( ) ( )2
cos x 1 cos2x m cos x m sin x *+ − =
a/ Giaûi (*) khi m= -2
b/ Tìm m sao cho (*) coù ñuùng hai nghieäm treân
2
0,
3
π⎡ ⎤
⎢ ⎥⎣ ⎦
( )( ) ( )
( ) ( )
( )( )
2 2
2
2
Ta coù (*) cos x 1 2cos x 1 m cos x m 1 cos x
cos x 1 2cos x 1 m cos x m 1 cos x 0
cos x 1 2cos x 1 m 0
⇔ + − − = −
⎡ ⎤⇔ + − − − −⎣ ⎦
⇔ + − − =
=
a/ Khi m = -2 thì (*) thaønh :
( )( )
( )
+ + =
⇔
⇔ = π + π ∈
π⎡ ⎤ ⎡
∈ = ∈⎢ ⎥ ⎢⎣ ⎦ ⎣
2
cos x 1 2 cos x 1 0
cosx = -1
x k2 k Z
2 1
b / Khi x 0, thì cos x t ,1
3 2
⎤
− ⎥⎦
Nhaän xeùt raèng vôùi moãi t treân
1
,1
2
⎡ ⎤
−⎢ ⎥⎣ ⎦
ta chæ tìm ñöôïc duy nhaát moät x treân
2
0,
3
π⎡ ⎤
⎢ ⎥⎣ ⎦
Yeâu caàu baøi toaùn coù ñuùng hai nghieäm treân2
2t 1 m 0⇔ − − =
1
,1
2
⎡ ⎤
−⎢ ⎥⎣ ⎦
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Xeùt ( ) ( )2
y 2t 1 P vaø y m d= − =
Ta coù y’ = 4t
Vaäy (*) coù ñuùng hai nghieäm treân
2
0,
3
π⎡ ⎤
⎢ ⎥⎣ ⎦
⇔ (d) caét (P) taïi hai ñieåm phaân bieät treân
1
,1
2
⎡ ⎤
−⎢ ⎥⎣ ⎦
⇔
1
1 m
2
− < ≤
Baøi 82 : Cho phöông trình ( ) ( )2 2
1 a tg x 1 3a 0 1
cos x
− − + + =
a/ Giaûi (1) khi
1
a
2
=
b/ Tìm a ñeå (1) coù nhieàu hôn moät nghieäm treân 0,
2
π⎛ ⎞
⎜ ⎟
⎝ ⎠
Ñieàu kieän : cos x 0 x k
2
π
≠ ⇔ ≠ + π
( ) ( ) ( )
( )( ) ( )
( ) ( )
( ) ( )
2 2
2 2
2
2
1 1 a sin x 2cos x 1 3a cos x 0
1 a 1 cos x 2cos x 1 3a cos x 0
4a cos x 2cos x 1 a 0
a 4 cos x 1 2cos x 1 0
2cos x 1 a 2cos x 1 1 0
⇔ − − + + =
⇔ − − − + + =
⇔ − + − =
⇔ − − − =
⇔ − + − =⎡ ⎤⎣ ⎦
a/ Khi
1
a
2
= thì (1) thaønh : ( )
1
2cos x 1 cos x 0
2
⎛ ⎞
− − =⎜ ⎟
⎝ ⎠
( )
( )
1
cos x cos nhaändo cos x 0
2 3
x k2 k Z
3
π
⇔ = = ≠
π
⇔ = ± + π ∈
b/ Khi x 0,
2
π⎛
∈ ⎜
⎝ ⎠
⎞
⎟ thì ( )cos x t 0,1= ∈
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Ta coù : (1)
( )
( )
1
cos x t 0,1
2
2a cos x 1 a 2
⎡
= = ∈⎢⇔
⎢
= −⎢⎣
Yeâu caàu baøi toaùn ⇔ (2) coù nghieäm treân ( )
a 0
1 1 a
0,1 0 1
2 2a
1 a 1
2a 2
⎧
⎪ ≠
⎪
−⎪⎧ ⎫
⇔ < <⎨ ⎬ ⎨
⎩ ⎭ ⎪
−⎪
≠⎪⎩
( )
a 0
0 a 11 a 1
0 a 1
12a 3
a 0 a
1 3a 13
0 a
12a 2
a
2 1 a 2a 2
≠⎧ ⎧
⎪ ⎪ < <− ⎧⎪ > < <⎪ ⎪⎪ ⎪ ⎪
⇔ ⇔ < ∨ > ⇔⎨ ⎨ ⎨−⎪ ⎪ ⎪< ≠
⎪⎪ ⎪ ⎩
≠⎪ ⎪− ≠ ⎩⎩
Caùch khaùc : daët u =
cos x
1
, ñieàu kieän u ; pt thaønh≥1
( ) ( )− − − + + = ⇔ − − +2 2
1 a ( u 1) 2u 1 3a 0 1 a u 2u 4a 0=
⇔ − − − =( u 2)[ (1 a)u 2a ] 0
Baøi 83 : Cho phöông trình : ( )cos4x 6sin x cos x m 1+ =
a/ Giaûi (1) khi m = 1
b/ Tìm m ñeå (1) coù hai nghieäm phaân bieät treân 0,
4
π⎡ ⎤
⎢ ⎥⎣ ⎦
Ta coù : (1) ⇔ − +2
1 2sin 2x 3sin 2x m=
( )
( )2
t sin2x t 1
2t 3t m 1 0 2
⎧ = ≤⎪
⇔ ⎨
− + − =⎪⎩
a/ Khi m = 1 thì (1) thaønh
( ) ( )
( )2
t sin 2x t 1
t sin 2x t 1
3
t 0 t loaïi2t 3t 0
2
k
sin 2x 0 x
2
⎧ = ≤⎧ = ≤⎪ ⎪
⇔⎨ ⎨
= ∨ =− =⎪ ⎪⎩
⎩
π
⇔ = ⇔ =
b/ Khi [ ]
π⎡ ⎤
∈ =⎢ ⎥⎣ ⎦
x 0, thì sin 2x t 0,1
4
∈
Nhaän thaáy raèng moãi t tìm ñöôïc treân [ ]0,1 ta chæ tìm ñöôïc duy nhaát moät
x 0,
4
π⎡ ⎤
∈ ⎢ ⎥⎣ ⎦
Ta coù : (2) ⇔ 2
2t 3t 1 m− + + =
Xeùt [ ]2
y 2t 3t 1treân 0,1= − + +
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Thì y ' 4t 3= − +
Yeâu caàu baøi toaùn ⇔ (d) y = m caét taïi hai ñieåm phaân bieät treân [ ]0,1
17
2 m
8
⇔ ≤ <
Caùch khaùc :ñaët . Vì a = 2 > 0, neân ta coù= − + −2
f(x) 2t 3t m 1
Yeâu caàu baøi toaùn ⇔
( )
( )
m
f m
f m
S
Δ = − >⎧
⎪ = − ≥⎪⎪
⎨ = − ≥
⎪
⎪ ≤ = ≤
⎪⎩
17 8 0
0 1 0
01 2
3
0 1
2 4
17
2 m
8
⇔ ≤ <
( )5 5 2
4 cos x.sin x 4sin x cos x sin 4x m 1− = +
a/ Bieát raèng laø nghieäm cuûa (1). Haõy giaûi (1) trong tröôøng hôïp ñoù.x = π
b/ Cho bieát x
8
π
= − laø moät nghieäm cuûa (1). Haõy tìm taát caû nghieäm cuûa (1) thoûa
4 2
x 3x 2− + < 0
( )
( )( )
( )
4 4 2
2 2 2 2 2
2
2
(1) 4sin x cos x cos x sin x sin 4x m
2sin 2x cos x sin x cos x sin x sin 4x m
2sin 2x.cos2x sin 4x m
sin 4x sin 4x m 0 1
⇔ − = +
⇔ − + =
⇔ = +
⇔ − + =
+
a/ laø nghieäm cuûa (1) = 0x = π 2
sin 4 sin4 m⇒ π − π +
m 0⇒ =
Luùc ñoù (1) ( )sin 4x 1 sin 4x 0⇔ − =
( )
⇔ = ∨ =
π
⇔ = π ∨ = + π
π π π
⇔ = ∨ = + ∈
sin 4x 0 sin 4x 1
4x k 4x k2
2
k k
x x k
4 8 2
Z
b/
2 2
4 2
2
t x 0 t x 0
x 3x 2 0
1 t 2t 3t 2 0
⎧ = ≥ ⎧ = ≥⎪
− + < ⇔ ⇔⎨ ⎨
< <− + <⎪ ⎩⎩
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( )
2
1 x 2 1 x 2
2 x 1 1 x 2 *
⇔ < < ⇔ < <
⇔ − < < − ∨ < <
π π⎛ ⎞
= − = − = −⎜ ⎟
⎝ ⎠
x thì sin 4x sin
8 2
1
( )x laø nghieämcuûa 1 1 1 m 0
8
m 2
π
= − ⇒ + + =
⇒ = −
Luùc ñoù (1) thaønh : 2
sin 4x sin4x 2 0− − =
( )
( )
( )
2
t sin 4x vôùi t 1
t t 2 0
t sin 4x vôùi t 1
t 1 t 2 loaïi
⎧ = ≤⎪
⇔ ⎨
− − =⎪⎩
⎧ = ≤⎪
⇔ ⎨
= − ∨ =⎪⎩
sin 4x 1
4x k2
2
k
x
8 2
⇔ = −
π
⇔ = − +
π π
⇔ = − +
π
Keát hôïp vôùi ñieàu kieän (*) suy ra k = 1
Vaäy (1) coù nghieäm
3
x
8 2 8
π π
= − + =
π
0thoûa 4 2
x 3x 2− + <
Baøi 85 : Tìm a ñeå hai phöông trình sau töông ñöông
( )
( )( ) (2
2cos x.cos2x 1 cos2x cos3x 1
4 cos x cos3x a cos x 4 a 1 cos2x 2
= + +
− = + − + )
( )
( )
2
Ta coù : (1) cos3x cos x 1 cos2x cos3x
cos x 1 2cos x 1
cos x 1 2cos x 0
1
cos x 0 cos x
2
⇔ + = + +
⇔ = + −
⇔ − =
⇔ = ∨ =
( ) ( )
( ) ( )
( )
⇔ − − = + −
⇔ + − − =
=⎡
⇔ ⎢
+ − + − =⎢⎣
2 3
3 2
2
Ta coù : (2) 4 cos x 4 cos x 3cos x a cos x 4 a 2 cos x
4 cos x 4 2a cos x a 3 cos x 0
cos x 0
4 cos x 2 2 a cos x a 3 0
2
[ ]⎛ ⎞
⇔ = − + − =⎜ ⎟
⎝ ⎠
1
cos x 0 hay cos x 2 cos x 3 a 0
2
−
⇔ = ∨ = ∨ =
1 a
cos x 0 cos x cos x
2 2
3
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Vaäy yeâu caàu baøi toaùn
a 3
0
2 a 3
a 3 1
a 4
2 2
a 1 a 5
a 3 a 3
1 1
2 2
−⎡
=⎢
=⎡⎢
− ⎢⎢⇔ = ⇔ =⎢⎢
⎢ < ∨ >⎢ ⎣− −⎢ < − ∨ >
⎢⎣
Baøi 86 : Cho phöông trình : cos4x = cos2
3x + asin2
x (*)
a/ Giaûi phöông trì nh khi a = 1
b/ Tìm a ñeå (*) coù nghieäm treân 0,
12
π⎛ ⎞
⎜ ⎟
⎝ ⎠
Ta coù : ( ) ( ) ( )
1 a
* cos4x 1 cos6x 1 cos2x
2 2
⇔ = + + −
( ) ( )
( )
( ) ( )
2 3
2 3
2 2cos 2x 1 1 4 cos 2x 3cos2x a 1 cos2x
t cos2x t 1
2 2t 1 1 4t 3t a 1 t
⇔ − = + − + −
⎧ = ≤⎪
⇔ ⎨
− = + − + −⎪⎩
( )
( )
( )
( )( ) ( ) ( )
3 2
2
t cos2x t 1
4t 4t 3t 3 a 1 t
1 cos2x t 1
t 1 4t 3 a 1 t * *
⎧ = ≤⎪
⇔ ⎨
− + + − = −⎪⎩
⎧ = ≤⎪
⇔ ⎨
− − + = −⎪⎩
a/ Khi a = 1 thì (*) thaønh :
( )
( )( )
( )
2
t cos2x t 1 t cos2x t 1
t 1 4t 4 0 t 1
⎧ = ≤ ⎧ = ≤⎪ ⎪
⇔⎨ ⎨
− − + = = ±⎪⎪ ⎩⎩
( )
⇔ = ± ⇔ =
π
⇔ = ⇔ = π ⇔ = ∈
2
cos 2x 1 cos 2x 1
k
sin 2x 0 2x k x , k Z
2
b/ Ta coù :
3
x 0, 2x 0, .Vaäy cos2x t ,1
12 6 2
⎛ ⎞π π⎛ ⎞ ⎛ ⎞
∈ ⇔ ∈ = ∈ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
( )( ) ( )
( )
⇔ − + = −
⇔ − = ≠
2
2
Vaäy (**) t-1 4t 3 a 1 t
4t 3 a do t 1
Xeùt ( )2 3
y 4t 3 P treân ,1
2
⎛ ⎞
= − ⎜ ⎟⎜ ⎟
⎝ ⎠
3
y ' 8t 0 t ,1
2
⎛ ⎞
⇒ = > ∀ ∈ ⎜ ⎟⎜ ⎟
⎝ ⎠
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Do ñoù (*) coù nghieäm treân ( ) ( )
⎛ ⎞π⎛ ⎞
⇔ = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
3
0, d : y a caét P treân ,1
2 2
( )
3
y a y
2
0 a 1
⎛ ⎞
⇔ < <⎜ ⎟⎜ ⎟
⎝ ⎠
⇔ < <
1
BAØI TAÄP
1. Giaûi caùc phöông trình sau :
a/ sin4x = tgx
b/ 4 4 4 9
sin x sin x x sin x
4 4
π π⎛ ⎞ ⎛ ⎞
+ + + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ 8
=
c/ tgx cot gx 4+ =
d/
( ) 2
sin x 3 2 2cos x 2sin x 1
1
1 sin 2x
− − −
=
−
e/ 4
4 cos x 3 2 sin 2x 8cos x+ =
f/
1 1 2
cos x sin2x sin4x
+ =
g/ sin 2x 2 sin x 1
4
π⎛ ⎞
+ −⎜ ⎟
⎝ ⎠
=
h/ ( ) ( )2 2sin x 1 4 sin x 1 cos 2x sin 2x
4 4
π π⎛ ⎞ ⎛ ⎞
− = − − + − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
k/ 24x
cos cos x
3
=
l/
x
tg .cos x sin2x 0
2
+ =
m/ 1 3tgx 2sin2x+ =
n/ cot gx tgx 2tg2x= +
p/ + =2 3x 4x
2cos 1 3cos
5 5
q/ 2
3cos4x 2cos 3x 1− =
r/ 2 3x
2cos 1 3cos2x
2
+ =
s/
x
cos x tg 1
2
+ =
t/ 2
3tg2x 4tg3x tg 3x.tg2x− =
u/ 2 3
cos x.cos4x cos2x.cos3x cos 4x
2
+ + =
v/ 2 2 2 2 3
cos x cos 2x cos 3x cos 4x
2
+ + + =
w/ sin4x tgx=
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x/ 6 6 213
cos x sin x cos 2x
8
+ =
y/
3 x 1 3x
sin sin
10 2 2 10 2
π π⎛ ⎞ ⎛
− = +⎜ ⎟ ⎜
⎝ ⎠ ⎝
⎞
⎟
⎠
2. ( 1 )6 6
sin x cos x a sin 2x+ =
a/ Giaûi phöông trình khi a = 1.
b/ Tìm a ñeå (1) coù nghieäm (ÑS :
1
a
4
≥ )
3. Cho phöông trình
( )
6 6
2 2
cos x sin x
2mtg2x 1
cos x sin x
+
=
−
a/ Giaûi phöông trình khi m =
1
8
b/ Tìm m sao cho (1) coù nghieäm (ÑS :
1
m
8
≥ )
4. Tìm m ñeå phöông trình
sin4x mtgx coù nghieäm x k= ≠ π
1
ÑS : m 4
2
⎛ ⎞
− < <⎜ ⎟
⎝ ⎠
5. Tìm m ñeå phöông trình :
cos3x cos2x mcos x 1 0− + − =
coù ñuùng 7 nghieäm treân ,2
2
π⎛ ⎞
− π⎜
⎝ ⎠
⎟ ( )ÑS :1 m 3< <
( ) ( )4 4 6 6 2
4 sin x cos x 4 sin x cos x sin 4x m+ − + − = coù nghieäm
1
ÑS : m 1
8
⎛ ⎞
− ≤ ≤⎜ ⎟
⎝ ⎠
7. Cho phöông trình :
(1)2 2 2
6sin x sin x m cos 2x− =
a/ Giaûi phöông trình khi m = 3
b/ Tìm m ñeå (1) coù nghieäm ( )ÑS :m 0≥
( )4 22m 1m
sin x cos4x sin4x sin x 0
4 4
+
+ + − =
coù hai nghieäm phaân bieät treân ,
4 2
π π⎛ ⎞
⎜ ⎟
⎝ ⎠
1
ÑS :2 5 4 m
2
⎛ ⎞
− < <⎜ ⎟
⎝ ⎠
coù nghieäm( )6 6 4 4
sin x cos x m sin x cos x+ = +
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1
ÑS : m 1
2
⎛ ⎞
≤ ≤⎜ ⎟
⎝ ⎠
2 2
cos4x cos 3x a sin x= +
Tìm a ñeå phöông trình coù nghieäm x 0,
2
π⎛ ⎞
∈ ⎜ ⎟
⎝ ⎠
( )ÑS :0 a 1< <
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CHÖÔNG IV
PHÖÔNG TRÌNH BAÄC NHAÁT THEO SIN VAØ COSIN
( PHÖÔNG TRÌNH COÅ ÑIEÅN )
( ) ( )a sin u bcosu c * . a, b R 0+ = ∈
Caùch 1 : Chia 2 veá phöông trình cho + ≠2 2
a b 0
Ñaët [ ]2 2 2 2
a b
vaø sin vôùi 0,2
a b a b
α = α = α ∈ π
+ +
cos
( )
( )
2 2
2 2
c
Thì * sin u cos cosu sin
a b
c
sin u
a b
⇔ α + α =
+
⇔ + α =
+
Caùch 2 :
Neáu u laø nghieäm cuûa (*) thì :k2= π + π
asin bcos c b cπ + π = ⇔ − =
Neáu u ñaëtk≠ π + π2
u
t tg
2
= thì (*) thaønh :
2
2 2
2t 1 t
a b
1 t 1 t
−
+ =
+ +
c
( ) ( )( )2
b c t 2at c b 0 1 vôùi b c 0⇔ + − + − = + ≠
Phöông trình coù nghieäm ( )( )2
' a c b c b 0⇔ Δ = − + − ≥
2 2 2 2 2 2
a c b a b c⇔ ≥ − ⇔ + ≥
Giaûi phöông trình (1) tìm ñöôïc t. Töø
u
t tg
2
= ta tìm ñöôïc u.
Baøi 87 : Tìm
2 6
x ,
5 7
π π⎛
∈ ⎜
⎝ ⎠
⎞
⎟ thoûa phöông trình : ( )cos7x 3 sin7x 2 *− = −
Chia hai veá cuûa (*) cho 2 ta ñöôïc :
( ) ⇔ − = −
π π
⇔ − + =
π π⎛ ⎞
⇔ − =⎜ ⎟
⎝ ⎠
1 3 2
* cos7x sin 7x
2 2 2
2
sin cos7x cos sin 7x
6 6
sin 7x sin
6 4
2
π π π π
⇔ − = + π − = +
3
7x k2 hay 7x h2
6 4 6 4
π , ( )∈k, h Z
π π π π
⇔ = + = + ∈
5 k2 11 h2
x hay x , k ,
84 7 84 7
h
Do
2 6
x ,
5 7
π π⎛
∈ ⎜
⎝ ⎠
⎞
⎟ neân ta phaûi coù :
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π π π π π π π π
< + < < + < ∈
2 5 k2 6 2 11 h2 6
hay ( k, h )
5 84 7 7 5 84 7 7
⇔ < + < < + < ∈
2 5 k2 6 2 11 h2 6
hay ( k, h )
5 84 7 7 5 84 7 7
Suy ra k = 2, =h 1,2
5 4 53 11 2 35
Vaäy x x
84 7 84 84 7 84
11 4 59
x
84 7 84
π π π π
= + = π ∨ = + =
π π
∨ = + = π
π
( )3
3sin 3x 3 cos9x 1 4sin 3x *− = +
Ta coù : ( ) ( )3
* 3sin 3x 4sin 3x 3 cos9x 1⇔ − − =
sin9x 3 cos9x 1⇔ − =
1 3
sin 9x cos9x
2 2
⇔ −
1
2
=
1
sin 9x sin
3 2
π π⎛ ⎞
⇔ − = =⎜ ⎟
⎝ ⎠ 6
π π π π
⇔ − = + π − = + π ∈
5
9x k2 hay 9x k2 , k
3 6 3 6
π π π π
⇔ = + = + ∈
k2 7 k2
x hay x ,
18 9 54 9
k
( )
1
tgx sin 2x cos2x 2 2cos x 0 *
cos x
⎛ ⎞
− − + − =⎜ ⎟
⎝ ⎠
Luùc ñoù : ( )
sin x 2
* sin2x cos2x 4cos x 0
cos x cos x
⇔ − − + − =
2
sin x sin 2x cos x cos x cos2x 4 cos x 2 0⇔ − − + − =
( )2
sin x 1 2cos x cos x cos2x 2cos2x 0⇔ − − + =
=
≠
sin xcos2x cos xcos2x 2cos2x 0⇔ − − + =
⇔ = − − +c os 2x 0 hay sin x cos x 2 0
( )
( )
⎡ = = − =
⎢⇔
⎢ + = + <⎢⎣
2
2 2 2
cos 2x 0 nhaän do cos 2x 2cos x 1 0 thì cos x 0
sin x cos x 2 voâ nghieäm vì 1 1 2
( )
π
⇔ = + ∈
π π
⇔ = + ∈
2x 2k 1 , k
2
k
x , k
4 2
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Baøi 90 : Giaûi phöông trình ( )
3 1
8sin x *
cos x sin x
= +
Luùc ñoù (*) 2
8sin xcosx 3 sin x cosx⇔ = +
( )
( )
⇔ − = +
⇔ − = −
⇔ − + = −
⇔ = − +
π⎛ ⎞
⇔ = +⎜ ⎟
⎝ ⎠
π π
⇔ = + + π ∨ = − − +
π π π
⇔ = + π ∨ = − + ∈
4 1 cos 2x cos x 3 sin x cos x
4 cos 2x cos x 3 sin x 3cos x
2 cos 3x cos x 3 sin x 3cos x
3 1
cos 3x sin x cosx
2 2
cos 3x cos x
3
3x x k2 3x x k2
3 3
k
x k x , k
6 12 2
π
Nhaän so vôùiñieàu kieän sin2x 0≠
Caùch khaùc :
(*) 2
8sin xcos x 3 sin x cos x⇔ = +
( hieån nhieân cosx = 0 hay sinx = 0 khoâng laø nghieäm cuûa pt naøy )
⇔ − = +2
8(1 cos x) cos x 3 sin x cos x
⇔ − = +3
8cos x 8cos x 3 sin x cos x
⇔ − = −3
6cos x 8cos x 3 sin x cos x
⇔ − = −3 1 3
4 cos x 3cos x cos x sin x
2 2
π⎛ ⎞
⇔ = +⎜ ⎟
⎝ ⎠
π π
⇔ = + + π ∨ = − − +
π π π
⇔ = + π ∨ = − + ∈
cos 3x cos x
3
3x x k2 3x x k2
3 3
k
x k x , k
6 12 2
π
( )9sin x 6cos x 3sin 2x cos2x 8 *+ − + =
Ta coù : (*) ( )2
9sin x 6cos x 6sin x cos x 1 2sin x 8⇔ + − + − =
( ) ( )
⇔ − − + −
⎛ ⎞
⇔ − − − −⎜ ⎟
⎝ ⎠
2
6cos x 6sin x cos x 2sin x 9sin x 7 0
7
6cos x 1 sin x 2 sin x 1 sin x 0
2
=
=
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( )
⎛ ⎞
⇔ − = + − =⎜ ⎟
⎝ ⎠
=⎡
⎢⇔
+ = + <⎢⎣
2 2 2
7
1 sin x 0 hay 6 cos x 2 sin x 0
2
sin x 1
6cos x 2sin x 7 voâ nghieäm do 6 2 7
π
⇔ = + π ∈x k2 , k
2
Baøi 92 : Giaûi phöông trình: ( )sin 2x 2cos2x 1 sin x 4 cos x *+ = + −
Ta coù : (*) ( )2
2sin x cos x 2 2cos x 1 1 sin x 4 cos x⇔ + − = + −
( )
⇔ − + + − =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⇔ − + − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⇔ − = + + = + <
2
2 2 2
2sin x cos x sin x 4 cos x 4 cos x 3 0
1 1 3
2sin x cos x 4 cos x cos x 0
2 2 2
1
cos x 0 hay 2sin x 4 cos x 6 0 voâ nghieäm do 2 4 6
2
π
⇔ = ± + πx k
3
2
( )2sin 2x cos2x 7sin x 2cos x 4 *− = + −
Ta coù : (*) ( )2
4sin x cos x 1 2sin x 7sin x 2cos x 4⇔ − − = + −
( )
( ) ( )
( ) ( )( )
( )
⇔ − + − + =
⎛ ⎞
⇔ − + − −⎜ ⎟
⎝ ⎠
⇔ − + − − =
⇔ − = + − = + <
2
2 2 2
2 cos x 2sin x 1 2sin x 7 sin x 3 0
1
2 cos x 2sin x 1 2 sin x sin x 3
2
2 cos x 2sin x 1 2sin x 1 sin x 3 0
2sin x 1 0 hay 2 cos x sin x 3 0 voâ nghieäm vì 1 2 3
π π
⇔ = + π ∨ = + π ∈
5
x k2 x k2 , k
6 6
( )sin 2x cos2x 3sin x cos x 2 *− = + −
Ta coù (*) ( )2
2sin x cos x 1 2sin x 3sin x cos x 2⇔ − − = + −
( )
( ) ( )(
⇔ − + − +
⇔ − + − −
⇔ − = + − =
2
cos x 2sin x 1 2sin x 3sin x 1 0
cos x 2sin x 1 sin x 1 2sin x 1 0
2sin x 1 0 hay cos x sin x 1 0
)
=
=
π⎛ ⎞
⇔ = −⎜ ⎟
⎝ ⎠
1
sin x hay 2 cos x x 1
2 4
=
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π π π π
⇔ = + π ∨ = + π − = ± + π ∈
5
x k2 x k2 hay x k2 , k
6 6 4 4
π π π
⇔ = + π ∨ = + π = + π ∨ = π ∈
5
x k2 x k2 hay x k2 x k2 , k
6 6 2
( ) ( )
2
sin 2x 3 cos2x 5 cos 2x *
6
π⎛ ⎞
+ − = −⎜ ⎟
⎝ ⎠
Ñaët t sin2x 3 cos2x= + , Điều kiện a b t a b− + = − ≤ ≤ = +2 2 2 2
2 2
Thì
1 3
t 2 sin 2x cos2x 2cos 2x
2 2
⎛ ⎞
6
π⎛ ⎞
= + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
−
Vaäy (*) thaønh:
− = ⇔ − − = ⇔ = ∨ = −2 2t 5
t 5 2t t 10 0 t ( loaïi ) t
2 2
2
Do ñoù ( )* ⇔ cos 2x 1
6
π⎛ ⎞
− = −⎜ ⎟
⎝ ⎠
π π
⇔ − = π + π ⇔ = +
7
2x k2 x k
6 1
π
2
Baøi 96 : Giaûi phöông trình ( )+ + =3
2cos x cos2x sinx 0 *
Ta coù (*) 3 2
2cos x 2cos x 1 sinx 0⇔ + − + =
( )
( )( ) ( )
( )( )
2
2
2cos x cosx 1 1 sinx 0
2 1 sin x 1 cosx 1 sinx 0
1 sinx 0 hay 2 1 sinx 1 cosx 1 0
⇔ + − + =
⇔ − + − − =
⇔ − = + + − =
2
1 sinx 0 hay 1 2sinxcosx 2(sinx cosx) 0
1 sinx 0 hay (sinx cosx) 2(sinx cosx) 0
⇔ − = + + + =
⇔ − = + + + =
( )2 2 2
sin x 1 haysin x cosx 0 hay sin x cosx 2 0 voâ nghieäm do: 1 1 2⇔ = + = + + = + <
sinx 1 haytgx 1⇔ = = − x k2 hay x k2 , k
2 4
π π
⇔ = + π = − + π ∈
1 cos2x
1 cotg2x *
sin 2x
−
+ =
Ñieàu kieän : sin2x 0 cos2x 1≠ ⇔ ≠ ±
Ta coù (*)
2
1 cos2x 1
1 cot g2x
1 cos2x1 cos 2x
1
cot g2x 1
1 cos2x
cos2x cos2x
sin2x 1 cos2x
−
⇔ + = =
+−
⇔ = −
+
−
⇔ =
+
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( )= ≠ ±⎡
⎢⇔ −⎢ =
⎢ +⎣
⇔ = ∨ + = −
⇔ = ∨ + =
cos2x 0 nhaän do 1
1 1
sin2x 1 cos2x
cos2x 0 1 cos2x sin2x
cos2x 0 sin2x cos2x 1−
1
cos2x 0 sin 2x sin
4 42
5
2x k 2x k2 2x k2 ,k
2 4 4 4 4
π π⎛ ⎞ ⎛ ⎞
⇔ = ∨ + = − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
π π π π π
⇔ = + π ∨ + = − + π ∨ + = + π ∈
( )
k
x x k 2x k2 loaïi ,
4 2 4
k
x , k
4 2
π π π
⇔ = + ∨ == − + π ∨ = π + π ∈
π π
⇔ = + ∈
k
Baøi 98 : Giaûi phöông trình ( ) ( )4 4
4 sin x cos x 3 sin 4x 2 *+ + =
Ta coù : (*)
( )
22 2 2 2
4 sin x cos x 2sin xcos x 3sin4x 2⎡ ⎤⇔ + − +⎢ ⎥⎣ ⎦
=
⎡ ⎤
⇔ − + =⎢ ⎥⎣ ⎦
21
4 1 sin 2x 3 sin4x 2
2
⇔ + = −
⇔ + =
π π⎛ ⎞
⇔ − =⎜ ⎟
⎝ ⎠
π π
⇔ − = ± + π
cos4x 3sin4x 1
1 3 1
cos4x sin4x
2 2
2
cos 4x cos
3 3
2
4x k2
3 3
−
2
4x k2 hay 4x k2 ,k
3
x k hay x k ,k
4 2 12 2
π
⇔ = π + π = − + π ∈
π π π π
⇔ = + = − + ∈
Caùch khaùc :
( )(*) 2
2 1 sin 2x 3 sin 4x 0⇔ − + =
2
2cos 2x 2 3sin2xcos2x 0
cos2x 0 cos2x 3sin2x 0
cos2x 0 cot g2x 3
⇔ + =
⇔ = ∨ +
⇔ = ∨ = −
=
2x k 2x k , k
2 6
k k
x x , k
4 2 12 2
π π
⇔ = + π ∨ = − + π ∈
π π π π
⇔ = + ∨ = − + ∈
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Baøi 99 : Giaûi phöông trình ( )3 3 1
1 sin 2x cos 2x sin4x *
2
+ + =
Ta coù (*) ( )( )
1
1 sin2x cos2x 1 sin2xcos2x sin4x
2
⇔ + + − =
( )
1 1
1 sin4x sin2x cos2x 1 sin4x 0
2 2
1
1 sin4x 0 hay 1 sin2x cos2x 0
2
⎛ ⎞
⇔ − + + − =⎜ ⎟
⎝ ⎠
⇔ − = + + =
( )sin4x 2 loaïi
sin2x cos2x 1
2 sin(2x ) 1
4
=⎡
⇔ ⎢
+ =⎣
π
−
⇔ + = −
( )
sin 2x sin( )
4 4
2x k2
4 4 k Z
5
2x k2
4 4
x k x k , k
4 2
π π⎛ ⎞
⇔ + = −⎜ ⎟
⎝ ⎠
π π⎡
+ = − + π⎢
⇔ ∈⎢
π π⎢ + = + π
⎢⎣
π π
⇔ = − + π ∨ = + π ∈
( )( )tgx 3cot gx 4 sin x 3 cosx *− = +
Ñieàu kieän
sinx 0
sin2x 0
cosx 0
≠⎧
⇔ ≠⎨
≠⎩
Luùc ñoù : (*) ( )sinx cosx
3 4 sinx 3 cos
cosx sinx
⇔ − = + x
( )
( )( )
2 2
sin x 3cos x 4sinxcosx sinx 3 cosx
sinx 3 cosx sinx 3 cosx 2sin2x 0
sinx 3 cosx
1 3
sinx cosx sin2x
2 2
⇔ − = +
⇔ + − − =
⎡ = −
⎢
⇔ ⎢
− =⎢⎣
tgx 3 tg
3
sin x sin2x
3
x k x 2x k2 x 2x k2 , k
3 3 3
⎡ π⎛ ⎞
= − = −⎜ ⎟⎢
⎝ ⎠⎢⇔
⎢ π⎛ ⎞
− =⎢ ⎜ ⎟
⎝ ⎠⎣
π π π
⇔ = − + π∨ − = + π ∨ − = π − + π ∈ Z
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( )
4 k2
x k x k2 x ,k
3 3 9 3
4 k2
x k x nhaän do sin2x 0
3 9 3
π π π π
⇔ = − + π ∨ = − − π ∨ = + ∈
π π π
⇔ = − + π ∨ = + ≠
Baøi 101 : Giaûi phöông trình ( )3 3
sin x cos x sin x cosx *+ = −
Ta coù : (*) 3 3
sin x sinx cos x cosx 0⇔ − + + =
( )
( )
( )
2 3
2 3
2
sinx sin x 1 cos x cosx 0
sinxcos x cos x cosx 0
cosx 0 hay sinxcosx cos x 1 0
cosx 0
sin2x cos2x 3 voâ nghieäm do 1 1 9
x 2k 1 , k Z
2
⇔ − + + =
⇔ − + + =
⇔ = − + + =
=⎡
⇔ ⎢
− + = − + <⎣
π
⇔ = + ∈
cos x sin x *
4 4
π⎛ ⎞
+ + =⎜ ⎟
⎝ ⎠
Ta coù : (*) ( )
2
21 1
1 cos2x 1 cos 2x
4 4 2
⎡ π ⎤⎛ ⎞ 1
4
⇔ + + − +⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
=
( ) ( )
2 2
1 cos2x 1 sin2x 1
cos2x sin2x 1
1 3
cos 2x cos
4 42
3
2x k2
4 4
x k x k , k
2 4
⇔ + + + =
⇔ + = −
π π⎛ ⎞
⇔ − = − =⎜ ⎟
⎝ ⎠
π π
⇔ − = ± + π
π π
⇔ = + π ∨ = − + π ∈ Z
4sin x.cos3x 4cos x.sin3x 3 3 cos4x 3 *+ + =
Ta coù : (*)
( ) ( )⇔ − + − +3 3 3 3
4sin x 4cos x 3cosx 4cos x 3sinx 4sin x 3 3 cos4x 3=
( )
⇔ − + + =
⇔ − + +
3 3
2 2
12sin xcosx 12sinxcos x 3 3 cos4x 3
4sinxcosx sin x cos x 3 cos4x 1=
2sin2x.cos2x 3 cos4x 1
sin
3sin4x cos4x 1
cos
3
⇔ +
π
⇔ + =
π
=
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sin4x.cos sin cos4x cos
3 3
π π
⇔ + =
3
π
sin 4x sin
3 6
5
4x k2 4x k2 , k
3 6 3 6
k k
x x , k
24 2 8 2
π π⎛ ⎞
⇔ + =⎜ ⎟
⎝ ⎠
π π π π
⇔ + = + π ∨ + = + π ∈
π π π π
⇔ = − + ∨ = + ∈
Baøi 104 : Cho phöông trình : ( )2 2
2sin x sin x cosx cos x m *− − =
a/ Tìm m sao cho phöông trình coù nghieäm
b/ Giaûi phöông trình khi m = -1
Ta coù : (*) ( ) ( )
1 1
1 cos2x sin2x 1 cos2x m
2 2
⇔ − − − + =
sin2x 3cos2x 2m 1⇔ + = − +
2
a/ (*) coù nghieäm 2 2
a b c⇔ + ≥
( )
2
2
1 9 1 2m
4m 4m 9 0
1 10 1 10
m
2 2
⇔ + ≥ −
⇔ − − ≤
− +
⇔ ≤ ≤
b/ Khi m = -1 ta ñöôïc phöông trình
( )sin2x 3cos2x 3 1+ =
( )
π
• = + = =Neáu x 2k 1 thì sin2x 0 vaø cos2x 1
2
− neân phöông trình (1) khoâng
thoûa.
( )
π
• ≠ + ≠ =Neáux 2k 1 thì cosx 0,ñaët t tgx
2
(1) thaønh
( )2
2 2
3 1 t2t
3
1 t 1 t
−
+ =
+ +
( ) (2 2
2
2t 3 1 t 3 t 1
6t 2t 0
t 0 t 3
⇔ + − = +
⇔ − =
⇔ = ∨ =
)
Vaäy (1) ⇔ tgx 0 hay tgx 3 tg x k= = = ϕ ⇔ = π hay x k , k= ϕ + π ∈
Baøi 105 : Cho phöông trình ( )2
3
5 4sin x
6tg2
*
sinx 1 tg
π⎛ ⎞
+ −⎜ ⎟ α⎝ ⎠ =
+ α
a/ Giaûi phöông trình khi
4
π
α = −
b/ Tìm α ñeå phöông trình (*) coù nghieäm
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Ta coù :
3
sin x sin x cosx
2 2
π π⎛ ⎞ ⎛ ⎞
− = − − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2
2
6tg 6sin
.cos 3sin2
1 tg cos
α α
= α = α vôùi cos 0
+ α α
α ≠
Vaäy : ( ) ( )
5 4cosx
* 3sin2 ñieàu kieän sinx 0 vaø cos 0
sinx
−
⇔ = α ≠ α ≠
3sin2 sinx 4cosx 5⇔ α + =
a/ Khi
4
π
α = − ta ñöôïc phöông trình
( )3sin x 4cosx 5 1− + = ( Hieån nhieân sin x = 0 khoâng laø nghieäm cuûa (1))
3 4
sinx cosx 1
5 5
⇔ − + =
Ñaët
3 4
cos vaø sin vôùi 0 2
5 5
ϕ = − ϕ = < ϕ < π
Ta coù pt (1) thaønh :
( )sin x 1ϕ + =
x k2
2
x k
2
π
⇔ ϕ+ = + π
π
⇔ = −ϕ+ + π2
≠b/ (**) coù nghieäm ( )
2
3sin2 16 25 vaø cos 0⇔ α + ≥ α
2
2
sin 2 1 vaø cos 0
sin 2 1
cos2 0
k
,k
4 2
⇔ α ≥ α ≠
⇔ α =
⇔ α =
π π
⇔ α = + ∈
BAØI TAÄP
1. Giaûi caùc phöông trình sau :
a/ ( )2 2 sinx cosx cosx 3 cos2x+ = +
b/ ( )(2cosx 1 sin x cosx 1− + ) =
c/ ( )2cos2x 6 cosx sinx= −
d/ 3sinx 3 3 cosx= −
e/ 2cos3x 3sinx cosx 0+ + =
f/ cosx 3 sin x sin2x cosx sinx+ = + +
g/
3
cosx 3sinx
cosx 3sinx 1
+ =
+ +
h/ sinx cosx cos2x+ =
k/ 3
4sin x 1 3sinx 3 cos3x− = −
i /
6
3cosx 4sinx 6
3cosx 4sinx 1
+ + =
+ +
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j/ cos7xcos5x 3sin2x 1 sin7xsin5x− = −
m/ ( )4 4
4 cos x sin x 3 sin 4x 2+ + =
p/ 2 2
cos x 3sin2x 1 sin x− = +
q/ ( )4sin2x 3cos2x 3 4sin x 1− = −
r/
2
tgx sin2x cos2x 4cosx
cosx
− − = − +
s/
( ) 2 x
2 3 cosx 2sin
2 4
1
2cosx 1
π⎛ ⎞
− − −⎜ ⎟
⎝ ⎠ =
−
2. Cho phöông trình cosx + msinx = 2 (1)
a/ Giaûi phöông trình m 3=
b/ Tìm caùc giaù trò m ñeå (1) coù nghieäm (ÑS : m 3≥ )
( )
msinx 2 m cosx 2
1
m 2cosx m 2sinx
− −
=
− −
a/ Giaûi phöông trình (1) khi m = 1
b/ Khi m 0 vaø m 2≠ ≠ thì (1) coù bao nhieâu nghieäm treân [ ]π π20 ,30 ?
(ÑS : 10 nghieäm)
4. Cho phöông trình
( )
2sinx cosx 1
a 1
sinx 2cosx 3
+ +
=
− +
a/ Giaûi (1)khi
1
a
3
=
b/ Tìm a ñeå (1) coù nghieäm
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CHÖÔNGV.
PHÖÔNG TRÌNH ÑOÁI XÖÙNG THEO SINX, COSX
( ) ( )a sin x cos x bsin x cos x c 1+ + =
Caùch giaûi
Ñaët = + ≤sin x cos x vôùi ñieàu kieän t 2t
Thì t 2 sin x 2 cos x
4 4
π π⎛ ⎞ ⎛ ⎞
= + = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Ta coù : ( )2
t 1 2sin x cos x neân 1 thaønh= +
( )2b
at t 1 c
2
+ − =
2
bt 2at b 2c 0⇔ + − − =
Giaûi (2) tìm ñöôïc t, roài so vôùi ñieàu kieän t 2≤
giaûi phöông trình
π⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
2 sin x t
4
ta tìm ñöôïc x
sin x sin x cos x 0 *+ + =
(*) ( ) ( )2
sin x 1 sin x cos x 1 sin x 0⇔ + + − =
( ) ( )⇔ + = + − =1 sin x 0 hay sin x cos x 1 sin x 0
( )
( )
sin x 1 1
sin x cos x sin x cos x 0 2
= −⎡
⇔ ⎢
+ − =⎢⎣
( ) ( )
( )
2
1 x k2 k Z
2
Xeùt 2 : ñaët t sin x cos x 2 cos x
4
ñieàu kieän t 2 thì t 1 2sin x cos x
π
• ⇔ = − + π ∈
π⎛ ⎞
• = + = −⎜ ⎟
⎝ ⎠
≤ = +
Vaäy (2) thaønh
2
t 1
t 0
2
−
− =
( )
2
t 2t 1 0
t 1 2
t 1 2 loaïi
⇔ − − =
⎡ = −
⇔ ⎢
= +⎢⎣
Do ñoù ( 2 ) ⇔ 2 cos x 1 2
4
π⎛ ⎞
− = −⎜ ⎟
⎝ ⎠
π⎛ ⎞
⇔ − = − = ϕ < ϕ < π⎜ ⎟
⎝ ⎠
π
⇔ − = ±ϕ + π ∈ ϕ = −
π
⇔ = ± ϕ + π ∈ ϕ = −
2
cos x 1 cos vôùi 0 2
4 2
2
x h2 , h , vôùi cos
4 2
2
x h2 , h , vôùi cos
4 2
1
1
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1 sin x cos x sin 2x *
2
− + + =
( ) ( )( )
3
* 1 sin x cos x 1 sin x cos x sin 2x
2
⇔ − + + − =
Ñaët t sin x cos x 2 sin x
4
π⎛ ⎞
= + = +⎜ ⎟
⎝ ⎠
Vôùi ñieàu kieän t 2≤
Thì 2
t 1 2sin xcos= + x
Vaäy (*) thaønh : ( )
2
2t 1 3
1 t 1 t 1
2 2
⎛ ⎞−
− + − = −⎜ ⎟⎜ ⎟
⎝ ⎠
( ) ( )
( )( )
( )
2 2
3 2
2
2 t 3 t 3 t 1
t 3t 3t 1 0
t 1 t 4t 1 0
t 1 t 2 3 t 2 3 loaïi
⇔ − + − = −
⇔ + − − =
⇔ − + + =
⇔ = ∨ = − + ∨ = − −
vôùi t = 1 thì
1
sin x sin
4 42
π π⎛ ⎞
+ = =⎜ ⎟
⎝ ⎠
π π π π
⇔ + = = π ∨ + = + π ∈
π
⇔ = π ∨ = + π ∈
3
x k2 x k2 , k
4 4 4 4
x k2 x k2 , k
2
vôùi
π −⎛ ⎞
= − + = =⎜ ⎟
⎝ ⎠
3 2
t 3 2 thì sin x sin
4 2
ϕ
π π −
⇔ + = ϕ + π ∨ + = π − ϕ + π ∈ =
π π −
⇔ = ϕ − + π ∨ = − ϕ + π ∈ = ϕ
3 2
x m2 x m2 , m , vôùi s
4 4 2
3 3
x m2 x m2 , m , vôùi sin
4 4 2
ϕin
2
Baøi 108 :Giaûi phöông trình ( ) ( )2 sin x cos x tgx cot gx *+ = +
Ñieàu kieän
sin x 0
sin 2x 0
cos x 0
≠⎧
⇔ ≠⎨
≠⎩
Luùc ñoù (*) ( )
sin x cos x
2 sin x cos x
cos x sin x
⇔ + = +
( )
2 2
sin x cos x 1
2 sin x cos x
sin x cos x sin x cos x
+
⇔ + = =
Ñaët t sin x cos x 2 sin x
4
π⎛ ⎞
= + = +⎜ ⎟
⎝ ⎠
Thì = + ≤ ≠2 2
t 1 2sin x cos x vôùi t 2 vaø t 1
(*) thaønh 2
2
2t
t 1
=
−
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3
2t 2t 2 0⇔ − − =
(Hieån nhieân khoâng laø nghieäm)t = ±1
( )( )
( )
2
2
t 2 2t 2t 2 0
t 2
t 2t 1 0 voâ nghieäm
⇔ − + + =
⎡ =
⇔ ⎢
+ + =⎢⎣
Vaäy ( ) ⇔* 2 sin x 2
4
π⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
π⎛ ⎞
⇔ + =⎜ ⎟
⎝ ⎠
π π
⇔ + = + π ∈
π
⇔ = + π ∈
sin x 1
4
x k2 , k
4 2
x k2 , k
4
Baøi 109 : Giaûi phöông trình ( ) ( ) (3 cot gx cos x 5 tgx sin x 2 *− − − = )
Vôùi ñieàu kieän sin , nhaân 2 veá phöông trình cho sinxcosx thì :2x 0≠ 0≠
( ) ( ) ( )⇔ − − − =2 2
* 3cos x 1 sin x 5sin x 1 cos x 2sin x cos x
( ) ( )
( ) ( )
( ) ( )
( )
( )
⇔ − − − = −
⇔ − + − − +⎡ ⎤ ⎡⎣ ⎦ ⎣
⇔ − + − − +
+ − =⎡
⇔ ⎢
− =⎢⎣
2 2
3cos x 1 sin x 5sin x 1 cos x 5sin x cos x 3sin x cos x
3cos x cos x 1 sin x sin x 5sin x sin x 1 cos x cos x 0
3cos x cos x sin x cos x sin x 5sin x sin x sin x cos x cos x 0
sin x cos x sin x cos x 0 1
3cos x 5sin x 0 2
=⎤⎦
=
( Ghi chuù: A.B + A.C = A.D ⇔ A = 0 hay B + C = D )
Giaûi (1) Ñaët t sin x cos x 2 sin x
4
π⎛ ⎞
= + = +⎜ ⎟
⎝ ⎠
Thì vôùi ñieàu kieän :2
t 1 2sin xcos= + x t 2 vaø t 1≤ ≠ ±
(1) thaønh :
2
2t 1
t 0 t 2t
2
−
− = ⇔ − − 1 0=
( )
( )
t 1 2 loaïi do t 2
t 1 2 nhaän so vôùi ñieàu kieän
⎡ = + ≤
⎢⇔
⎢ = −⎣
Vaäy ( )
1 2
sin x sin 0 2
4 2
π −⎛ ⎞
+ = = α < α <⎜ ⎟
⎝ ⎠
π
π π⎡ ⎡
+ = α + π = α − + π⎢ ⎢
⇔ ⇔⎢ ⎢
π π⎢ ⎢+ = π − α + π ∈ = − α + π ∈
⎢ ⎢⎣ ⎣
x k2 x k2
4 4
3
x k2 , k x k2 ,
4 4
k
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( ) ( )⇔ = = β ⇔ = β + π ∈ < β < π
3
2 tgx tg x h , h vôùi 0
5
( )
( )3 2
2
3 1 sinx x
3tg x tgx 8cos *
4 2cos x
+ π⎛ ⎞
− + = −⎜ ⎟
⎝ ⎠
Ñieàu kieän : cosx 0 sinx 1≠ ⇔ ≠ ±
Luùc ñoù : (*) ( ) ( )( )2 2
tgx 3tg x 1 3 1 sinx 1 tg x 4 1 cos x
2
⎡ ⎤π⎛ ⎞
⇔ − + + + = + −⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
( )4 1 sinx= +
( ) ( ) ( )
( )( )
( )( )
( )
( )
2 2
2
2
2
tgx 3tg x 1 1 sinx 3 1 tg x 4 0
3tg x 1 tgx 1 sinx 0
3tg x 1 sinx cosx sinxcosx 0
3tg x 1 1
sinx cosx sinxcosx 0 2
⎡ ⎤⇔ − + + + −⎣ ⎦
⇔ − + + =
⇔ − + + =
⎡ =
⇔ ⎢
+ + =⎢⎣
=
( )
2 1 3
(1) tg x tgx x k
3 3 6
Giaûi 2 ñaët t sinx cosx 2 sin x
4
π
• ⇔ = ⇔ = ± ⇔ = ± + π
π⎛ ⎞
• = + = ⎜ ⎟
⎝ ⎠
+
Vôùi ñieàu kieän t 2 vaø t≤ ≠ 1±
Thì 2
t 1 2sinxcosx= +
(2) thaønh :
2
2t 1
t 0 t 2t 1
2
−
+ = ⇔ + − = 0
( )
( )
t 1 2 loaïi doñieàu kieän t 2
t 1 2 nhaän so vôùi ñieàu kieän
⎡ = − − ≤
⎢⇔
⎢ = − +⎣
Vaäy
2 1
sin x sin
4 2
π −⎛ ⎞
+ = =⎜ ⎟
⎝ ⎠
ϕ
x k2 ,k x k2 ,k
4 4
3
x k2 ,k x k2
4 4
π π⎡ ⎡
+ = ϕ + π ∈ = ϕ − + π ∈⎢ ⎢
⇔ ⇔⎢ ⎢
π π⎢ ⎢+ = π − ϕ + π ∈ = − ϕ + π ∈
⎢ ⎢⎣ ⎣
,k
Baøi 111 : Giaûi phöông trình ( )− = − +3 3
2sin x sinx 2cos x cosx cos2x *
( ) ( ) ( )3 3 2 2
* 2 sin x cos x sin x cosx sin x cos x 0⇔ − − − + − =
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( ) ( )
( )
( )
sinx cosx 0 hay 2 1 sinxcosx 1 sinx cosx 0
sinx cosx 0 1
sinx cosx sin2x 1 0 2
⇔ − = + − + + =
− =⎡
⇔ ⎢
+ + + =⎢⎣
( )
( )
1 tgx 1
x k ,k
4
xeùt 2 ñaët t sinx cosx 2 cosx x
4
• ⇔ =
π
⇔ = + π ∈
π⎛ ⎞
• = + = ⎜ ⎟
⎝ ⎠
−
Vôùi ñieàu kieän : t 2≤
2
t 1 sin2= + x
( ) ( )2
Vaäy 2 thaønh t t 1 1 0+ − + =
( )t t 1 0 t 0 t 1⇔ + = ⇔ = ∨ = −
Khi t = 0 thì cos x 0
4
π⎛ ⎞
− =⎜ ⎟
⎝ ⎠
( )x 2k 1 ,k
4 2
3
x k ,k
4
π π
⇔ − = + ∈
π
⇔ = + π ∈
Khi
1 3
t 1 thì cos x cos
4 42
π π⎛ ⎞
= − − = − =⎜ ⎟
⎝ ⎠
3
x k2 ,k
4 4
x k2 hay x k2 ,k
2
π π
⇔ − = ± + π ∈
π
⇔ = π + π =− + π ∈
( )2 3 4 2 3 4
sinx sin x sin x sin x cosx cos x cos x cos x *+ + + = + + +
Ta coù : (*)
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 2 3 3 4 4
sinx cosx sin x cos x sin x cos x sin x cos x 0
sinx cosx 0 hay 1 sinx cosx 1 sinx.cosx sinx cosx 0
⇔ − + − + − + − =
⇔ − = + + + + + + =
( )
( ) ( )
sinx cosx 0 1
2 sinx cosx sinxcosx 2 0 2
− =⎡
⇔ ⎢
+ + + =⎢⎣
Ta coù : (1) tgx 1⇔ =
x k ,k
4
π
⇔ = + π ∈
Xeùt (2) : ñaët t sin x cosx 2 cos x
4
π⎛ ⎞
= + = −⎜ ⎟
⎝ ⎠
Vôùi ñieàu kieän t 2≤
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Thì 2
t 1 2sinxcosx= +
(2) thaønh
2
t 1
2t 2 0
2
−
+ + =
( )
2
t 4t 3 0
t 1 t 3 loaïi
⇔ + + =
⇔ = − ∨ = −
khi t = -1 thì
1 3
cos x cos
4 42
π π⎛ ⎞
− = − =⎜ ⎟
⎝ ⎠
3
x k2 ,k
4 4
3
x k2 ,
4 4
x k2 ,k
x k2 ,k
2
π π⎡
− = + π ∈⎢
⇔ ⎢
π π⎢ − = − + π ∈
⎢⎣
= π + π ∈⎡
⎢⇔ π⎢ = − + π ∈
⎣
k
Baøi 113 : Giaûi phöông trình ( ) ( )− + − =2 3 3
tg x 1 sin x cos x 1 0 *
Ñieàu kieän : cosx 0 sinx 1≠ ⇔ ≠ ±
Luùc ñoù (*) ( )
2
3 3
2
sin x
1 sin x cos x 1 0
cos x
⇔ − + − =
( )( ) ( )( )
( )( )
( )( ) ( )( )
2 3 3 2
2 2
1 cos x 1 sin x 1 cos x 1 sin x 0
1 cosx 1 sinx 0
hay 1 cosx 1 sinx sin x 1 cosx cos x 1 sinx 0
⇔ − − − − − =
⇔ − − =
+ + + − + + +
( )
( )
2 2 2 2
cosx 1 nhaän do ñieàu kieän
sinx 1 loaïido ñieàu kieän
sin x sin xcosx cos x sinxcos x 0
⎡ =
⎢
⇔ =⎢
⎢
+ − − =⎢⎣
=
( )2 2
cosx 1
sin x cos x sinxcosx sinx cosx 0
=⎡
⇔ ⎢
− + − =⎣
cosx 1
sinx cosx 0 hay sinx cosx sinxcosx 0
=⎡
⇔ ⎢ − = + + =⎣
cosx 1 tgx 1
sinx cosx sinxcosx 0
= ∨ =⎡
⇔ ⎢ + + =⎣
x k2 ,k
x k ,k
4
sin x cosx sin xcosx 0
= π ∈⎡
⎢ π⎢⇔ = + π ∈
⎢
⎢ + + =⎣
xeùt pt sinx cosx sinxcosx 0+ + =
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ñaët
( )t sinx cosx 2 cosx x ñieàu kieän t 2 vaø t 1
4
π⎛ ⎞
= + = − ≤ ≠ ±⎜ ⎟
⎝ ⎠
2
t 1 2sin x cosx⇒ = +
Ta ñöôïc phöông trình
2
2t 1
t 0 t 2t
2
−
1 0+ = ⇔ + − =
( )
( )
t 1 2 loaïi
t 1 2 nhaän so vôùi ñk
⎡ = − −
⎢⇔
⎢ = − +⎣
Vaäy
2 1
cos x cos
4 2
π −⎛ ⎞
− = = ϕ⎜ ⎟
⎝ ⎠
x k2 ,k x k2 ,
4 4
π π
⇔ − = ±ϕ + π ∈ ⇔ = ± ϕ + π ∈k
Baøi 114 : Cho phöông trình ( ) ( )m sinx cosx 1 1 sin2x *+ + = +
Tìm m ñeå phöông trình coù nghieäm thuoäc ñoaïn 0,
2
π⎡ ⎤
⎢ ⎥
⎣ ⎦
Ñaët t sinx cosx 2 sin x
4
π⎛
= + = −⎜
⎝ ⎠
⎞
⎟, ñieàu kieän t 2≤
Thì 2
t 1 sin2x= +
Vaäy (*) thaønh : ( ) 2
m t 1 t+ =
Neáu
3
0 x thì x
2 4 4 4
π π π π
≤ ≤ ≤ + ≤
Do ñoù
2
sin x 1
2 4
π⎛ ⎞
≤ + ≤⎜ ⎟
⎝ ⎠
1 t 2⇔ ≤ ≤
ta coù ( ) 2
m t 1 t+ =
2
t
m
t 1
⇔ =
+
(do t = -1 khoâng laø nghieäm cuûa phöông trình)
Xeùt
2
t
y treân 1,
t 1
⎡ ⎤= ⎣ ⎦+
2
Thì
( )
2
2
t 2t
y' 0 t 1, 2
t 1
+ ⎡ ⎤= > ∀ ∈⎣ ⎦+
Vaäy y taêng treân 1, 2⎡ ⎤
⎣ ⎦
Vaäy (*) coù nghieäm treân ( ) ( )1, y 1 m y 2
2
π⎡ ⎤
⇔ ≤ ≤⎢ ⎥
⎣ ⎦
( )⇔ ≤ ≤ −
1
m 2 2 1
2
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